datavorous/entrance-exam-dataset huggingface dataset:
A solid spherical ball of radius 59 m is connected to a point A on the wall with the help of a string which makes an angle Î¸=45Â° with the vertical. The sphere can rotate freely about its central axis and it is set into rotational motion against the vertical face of the wall with an angular velocity 100 rad s-1. In how much time (in s) will it come to rest? [Î¼=0.1]

Options:

The correct answer is:
20

Solution:
FBD of the spherical ball
For translational equilibrium
T cos45Â°=Mg+Î¼N
T sin15Â° = N
Also ÏÎt=ÎL
âÎ¼NR Ît=IÏ0
Ît=25MR2Ï0Î¼NR
Solving we get t = 20 s

Topic: Physics, Rotational Motion, JEE Main
A solid spherical ball rolls on a horizontal surface at 10 m s-1 and continues to roll up on an inclined surface as shown in the figure. If the mass of the ball is 11 kg and frictional losses are negligible, the value of h, where the ball stops and starts rolling down the inclination is (Assume g=10 m s-2)

Options:
A 8 m
B 6 m
C 7 m
D 10 m

The correct answer is:
7 m

Solution:
Given,

Initial velocity of solid spherical ball is v=10 m s-1

Mass of ball rolling on the surface is m=11 kg

Applying work energy theorem,

W=∆KE

Wg=KEf-KEi ...1

Now, initial kinetic energy of spherical ball is Ki=12mv21+k2R2

And final kinetic energy is given by

Kf=0

Work done due to gravity is

Wg=-mgh

Substitute the values in equation (1) we get

⇒ -mgh=0-12×11×1021+25

Where, k2R2=25

⇒ h=35050=7 m

Topic: Physics, Rotational Motion, JEE Main
A solid spherical conductor is given a charge. The electrostatic potential of the conductor is

Options:
A Largest at the centre
B Largest on the surface
C Largest somewhere between the centre and the surface
D Constant throughout the conductor

The correct answer is:
Constant throughout the conductor

Solution:
For solid sphre charge is distributed uniformly so, Electric potential is constant every where in the sphere
( equal to k q R , where k = 1 4 π ε 0 )
within or on the surface of conductor.

Topic: Physics, Electrostatics, JEE Main
A solid uniform sphere resting on a rough horizontal plane is given a horizontal impulse directed through its centre so that it starts sliding with an initial velocity $v_{0}$. When it finally atarta rolling without shpping the speed of its centre is

Options:
A $\frac{2}{7} v_{0}$
B $\frac{3}{7} v_{0}$
C $\frac{5}{7} v_{0}$
D $\frac{6}{7} v_{0}$

The correct answer is:
$\frac{5}{7} v_{0}$

Solution:
Let, the final velocity be V

So, angular momentum will remain conserved along point of contact By conservation of angular momentum Angular momentum will remain conserved along point of contact
$$
\begin{aligned}
I_{\omega} &=\text { constant } \\
m v_{0} r &=m v r+\frac{2}{5} m r^{2} \times \omega \quad\left(\because \omega=\frac{v}{r}\right) \\
m v_{0} r &=m v r+\frac{2}{5} m r^{2}\left(\frac{v}{r}\right) \\
v_{0} &=v+\frac{2}{5} v \\
v_{0} &=\frac{7}{5} v \Rightarrow v=\frac{5}{7} v_{0}
\end{aligned}
$$

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2014
A solid which is transparent to visible light and whose conductivity increases with temperature is formed by

Options:
A
Metallic binding

B
Ionic binding

C
Covalent binding

D
Van der Waals binding

The correct answer is:
Covalent binding

Solution:
Covalent binding

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2006
A solid XY has NaCl structure. If radius of X+ is 100 pm. What is the radius of Y- ion:

Options:
A 120 pm
B 136.6 to 241.6 pm
C 136.6âpm
D 241.6âpm

The correct answer is:
136.6 to 241.6 pm

Solution:
For NaCl like structure. The radius ratio should be 0.414 to 0.732
r+r-=100rmax-=0.414
100rmin-=0.732
â´rmax-=241.6 pm
â´ r min =136.6âpm

Topic: Chemistry, Solid State, JEE Main
A solid XY kept in an evacuated sealed container undergoes decomposition to form a mixture of gases X and Y at temperature T.

The equilibrium pressure is 10 bar in this vessel. Kp for this reaction is?

Options:
A 25
B 100
C 10
D 5

The correct answer is:
25

Solution:
XY⇌X(S)(g) +Y(g)
Peq=Px+Px=10
⇒Px=5
Kp=Px.(Py)
=5×5=25

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2016 (10 Apr Online)
A solute A dimerizes in water. The boiling point of a 2 molal solution of A is 100.52C∘. The percentage association of A is ___ .
(Round off to the Nearest integer)

Use : Kb for water =0.52 K kg mol-1

Boiling point of water =100°C

Options:

The correct answer is:
100

Solution:
ΔTb=Tb-Tb0

100.52-100

=0.52°C

i=1-α2

∵ΔTb=iKb×m

0.52=1-α2×0.52×2

α=1

So, percentage association =100%.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (18 Mar Shift 2)
A solution at 20oC is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are 74.7 torr and 22.3 torr respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively:

Options:
A 30.5 torr and 0.389
B 35.0 torr and 0.480
C 38.0 torr and 0.589
D 35.8 torr and 0.280

The correct answer is:
38.0 torr and 0.589

Solution:
XBenzene =1.55=0.3
XToluene=3.55=0.7
According to Raoult's Law

P=xAP°A +xBP°B

Ptotal=0.3×74.7+0.7×22.3
=22.41+15.61=38.02
≈38 Torr
By Dalton's law to vapour phase
Mole fraction of a gas in a mixture of gas = Partial pressure of gas P 1 Total no.moles in the mixture
XBenzene′Vapour phase=0.3×74.738=22.4138
=0.589

Topic: Chemistry, Solutions, JEE Main, JEE Main 2015 (10 Apr Online)
A solution containing 2 g of a non-volatile solute in 20 g of water boils at 373.52 K. The molecular mass of the solute is _____ gmol-1. (Nearest integer)

Given, water boils at 373 K,Kb for water =0.52K kg mol-1

Options:

The correct answer is:
100

Solution:
Elevation in boiling point:
⇒ΔTb=373.52-373=0.52

ΔTb=Kb·m

⇒0.52=0.52×2 Molar Mass of Solute ×120×10-3

Molar Mass of solute=100 g/mol

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (30 Jan Shift 1)
A solution containing 10âg of a non-volatile, higher nonelectrolyte and 400âg of water boils at 100 .256 o C at 1âatm . The molecular weight of the solute (in âg/mol) is
Given: (Kb for water 0.512 C Î¿/m)

Options:

The correct answer is:
50

Solution:
ÎTb=iÃKbÃm
0.256=1Ã0.512Ã10.0M0.4
M=0.512Ã10.00.4Ã0.256
M=50 g/mol

Topic: Chemistry, Solutions, JEE Main
A solution containing 10 g per \(\mathrm{dm}^3\) of urea (molecular mass \(=60 \mathrm{~g} \mathrm{~mol}^{-1}\)) is isotonic with a \(5 \%\) solution of a nonvolatile solute. The molecular mass of this nonvolatile solution is:

Options:
A \(250 \mathrm{~g} \mathrm{~mol}^{-1}\)
B \(300 \mathrm{~g} \mathrm{~mol}^{-1}\)
C \(350 \mathrm{~g} \mathrm{~mol}^{-1}\)
D \(200 \mathrm{~g} \mathrm{~mol}^{-1}\)

The correct answer is:
\(350 \mathrm{~g} \mathrm{~mol}^{-1}\)

Solution:
The molar concentration of urea solution
$$
=\frac{10 g / d m^3}{\text { Mol. wt. of urea }}=\frac{10}{60} \mathrm{M}=\frac{1}{6} \mathrm{M}
$$
Molar concentration of $5 \%$ nonvolatile solute
$$
\begin{aligned}
& =\frac{50 \mathrm{~g} / \mathrm{dm}^3}{\text { mol. wt. of non - volatile solute }} \\
& =\frac{50}{\mathrm{~m}} \mathrm{M}
\end{aligned}
$$
Both solutions are isotonic to each other;
therefore, $\quad \frac{1}{6}=\frac{50}{m}$
$$
m=50 \times 6=300 \mathrm{~g} \mathrm{~mol}^{-1}
$$
Caution
$10 \mathrm{~g}$ per $\mathrm{dm}^3$ of urea is isotonic with $5 \%$ solution of a non-volatile solute. Hence, osmosis cannot occur between the two solutions. Thus, their molar concentrations are equal to each other.

Topic: Chemistry, Solutions, JEE Main
A solution containing \(10 \mathrm{~g}\) of an electrolyte \(\mathrm{AB}_2\) in \(100 \mathrm{~g}\) of water boils at \(100.52^{\circ} \mathrm{C}\). The degree of ionization of the electrolyte \((\alpha)\) is _______ \(\times 10^{-1}\). (nearest integer)
[Given : Molar mass of \(\mathrm{AB}_2=200 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{~K}_{\mathrm{b}}\) (molal boiling point elevation const. of water) \(=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\), boiling point of water \(=100^{\circ} \mathrm{C} ; \mathrm{AB}_2\) ionises as \(\left.\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2 \mathrm{~B}^{-}\right]\)

Options:

The correct answer is:
5

Solution:
$\begin{aligned} & \mathrm{AB}_2 \rightarrow \mathrm{A}^{+2}+2 \mathrm{~B}^{\ominus} \\ & \mathrm{i}=1+(3-1) \alpha \\ & \mathrm{i}=1+2 \alpha \\ & \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{k}_{\mathrm{b}} \mathrm{im} \\ & 0.52=0.52(1+2 \alpha) \frac{\frac{10}{200}}{\frac{100}{1000}} \\ & 1=(1+2 \alpha) \frac{10}{20} \\ & 2=1+2 \alpha \\ & \alpha=0.5\end{aligned}$
Ans. $\alpha=5 \times 10^{-1}$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (08 Apr Shift 1)
A solution containing 1.8 g of a compound (empirical formula CH2O) in 40 g of water is observed to freeze at -0.465âC. The molecular formula of the compound is-
(Kf of water=1.86 kg K mol-1)

Options:
A C2H4O2
B C3H6O3
C C5H10O5
D C6H12O6

The correct answer is:
C6H12O6

Solution:
M2=KfÃw2Ã1000â³TfÃw1
or M2=1.86Ã1.8Ã10000.465Ã40=180
n=180emp.formula mass = 18030=6
Molecular formula of the compound is C6H12O6 .

Topic: Chemistry, Solutions, JEE Main
A solution containing $0.85 \mathrm{~g}$ of $\mathrm{ZnCl}_2$ in $125.0 \mathrm{~g}$ of water freezes at $-0.23^{\circ} \mathrm{C}$. The apparent degree of dissociation of the salt is $\left(K_f\right.$ for water $=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, atomic mass: $\mathrm{Zn}=65.3$ and $\mathrm{Cl}=35.5)$

Options:
A
$1.36 \%$

B
$73.5 \%$

C
$7.35 \%$

D
$2.47 \%$

The correct answer is:
$73.5 \%$

Solution:
$$
\text { Mol.wt. }=\frac{k_f \times w \times 1000}{\Delta T_f \times W}=\frac{1.86 \times 0.85 \times 1000}{0.23 \times 125} \approx 55 \mathrm{gm}
$$
Where
$$
\begin{aligned}
& w=0.85 \mathrm{~g} \\
& W=125 \mathrm{~g} \\
& \Delta T_f=0^{\circ} \mathrm{C}-\left(-23^{\circ} \mathrm{C}\right)=23^{\circ} \mathrm{C}
\end{aligned}
$$
Now, $i=\frac{M_{\text {normal }}}{M_{\text {observed }}}=\frac{136.3}{55}=2.47$
$$
\begin{aligned}
& \underset{1-\alpha}{\mathrm{ZnCl}_2} \leftrightharpoons \underset{\alpha}{\mathrm{Zn}^{++}}+\underset{2 \alpha}{2 \mathrm{Cl}^{-}} \\
& \text {Van't Hoff factor ( } i \text { ) } \\
& =\frac{1-\alpha+\alpha+2 \alpha}{1}=2.47 \\
& \therefore \quad \alpha=0.735=73.5 \% \\
&
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2012 (12 May Online)
A solution containing $2.675 \mathrm{~g}$ of $\mathrm{CoCl}_3 .6 \mathrm{NH}_3$ (molar mass $=267.5 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $\mathrm{AgNO}_3$ to give $4.78 \mathrm{~g}$ of $\mathrm{AgCl}$ (molar mass $=143.5 \mathrm{~g} \mathrm{~mol}^{-1}$ ). The formula of the complex is (At. Mass of $\mathrm{Ag}=108 \mathrm{u}$ )

Options:
A
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$

B
$\left[\mathrm{CoCl}_2\left(\mathrm{NH}_3\right)_4\right] \mathrm{Cl}$

C
$\left[\mathrm{CoCl}_3\left(\mathrm{NH}_3\right)_3\right]$

D
$\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right] \mathrm{Cl}_2$

The correct answer is:
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$

Solution:
$\mathrm{CoCl}_3 .6 \mathrm{NH}_3 \rightarrow \times \mathrm{Cl}^{-} \stackrel{\mathrm{AgNO}_3}{\longrightarrow} \times \mathrm{AgCl} \downarrow$
$\mathrm{n}(\mathrm{AgCl})=\mathrm{x}\left(\mathrm{CoCl}_3 .6 \mathrm{NH}_3\right)$
$\frac{4.78}{143.5}=\times \frac{2.675}{267.5} \quad \therefore \mathrm{x}=3$
$\therefore$ The complex is $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2010
A solution containing a period-IV cation gives a precipitate on passing H2S. A solution of this precipitate in dil. HCl produces a white precipitate with NaOH solution and bluish-white precipitate with basic potassium ferrocyanide. The cation is

Options:
A Mn2+
B Zn2+
C Ni2+
D CO2+

The correct answer is:
Zn2+

Solution:

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2017 (08 Apr Online)
A solution containing active cobalt ${ }_{27}^{60} \mathrm{Co}$ having activity of $0.8 \mu \mathrm{Ci}$ and decay constant $\lambda$ is injected in an animal's body. If $1 \mathrm{~cm}^3$ of blood is drawn from the animal's body after $10 \mathrm{hrs}$ of injection, the activity found was 300 decays per minute. What is the volume of blood that is flowing in the body? $\left(1 \mathrm{Ci}=3.7 \times 10^{10}\right.$ decay per second and at $\left.\mathrm{t}=10 \mathrm{hrs}^{-\lambda \mathrm{t}}=0.84\right)$

Options:
A
6 litres

B
7 litres

C
4 litres

D
5 litres

The correct answer is:
5 litres

Solution:
Let initial activity $=\mathrm{No}=0.8 \mu \mathrm{ci}$ $0.8 \times 3.7 \times 10^4 \mathrm{dps}$
Activity in $1 \mathrm{~cm}^3$ of blood at $\mathrm{t}=10 \mathrm{hr}$, $\mathrm{n}=\frac{300}{60} \mathrm{dps}=5 \mathrm{dps}$
$\mathrm{N}=$ Activity of whole blood at time $\mathrm{t}=10 \mathrm{hr}$. Total volume of the blood in the person, V $=\frac{N}{n}$
$=\frac{N_0 e-\lambda t}{n}=\frac{0.8 \times 3.7 \times 10^4 \times 0.7927}{5} \cong 5$ litre

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A solution containing active cobalt Co2760 having an activity of 0.8 Î¼Ci and decay constant Î» is injected in an animal's body. When 1 cm3 blood is drawn from the animal's body after 10 h of injection, the activity was found to be 300 dpm(decays per minute), then the total volume of blood in the animal's body is close to [1 Ci=3.7Ã1010 dps(decays per second) and at t=10 h the value of e-Î»t=0.84]

Options:
A 4 L
B 6 L
C 5 L
D 7 L

The correct answer is:
5 L

Solution:
Let the total volume of blood is v, initial activity A0=0.Q μCi its activity at time t, A= A o e −λt , the activity of x volume, A1=AVX=XAoVe-λt
⇒V=XA0A1e-λt⇒V=1 cm38×10-7×3.7×101030060 0.84
=4.97×103 cm2=4.97 liter=5 L.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2018 (15 Apr)
A solution containing active cobalt Co2760 having an activity of 0.8 Î¼Ci and decay constant Î» is injected in an animal's body. When 1 cm3 blood is drawn from the animal's body after 10 h of injection, the activity was found to be 300 dpm(decays per minute), then the total volume of blood in the animal's body is close to [1 Ci=3.7Ã1010 dps(decays per second) and at t=10 h the value of e-Î»t=0.84]

Options:
A 4 L
B 6 L
C 5 L
D 7 L

The correct answer is:
5 L

Solution:
Let total volume of blood is v, initial activity A0=0.Q Î¼Ci its activity at time t
A= A o e âÎ»t
activity of x volume
A1=AVX=XAoVe-Î»t
V=XA0A1e-Î»t
V=1 cm38Ã10-7Ã3.7Ã101030060 0.84
=4.97Ã103cm2=4.97 litre

Topic: Physics, Nuclear Physics, JEE Main
A solution containing components $\mathrm{A}$ and B follows Raoult's law if:

Options:
A $\mathrm{A}-\mathrm{B}$ attraction force is greater than $\mathrm{A}-\mathrm{A}$ and $\mathrm{B}-\mathrm{B}$
B $\mathrm{A}-\mathrm{B}$ attraction force is less than $A-A$ and $B-B$.
C attraction force remains same as in $\mathrm{A}-\mathrm{A}$ and $\mathrm{B}-\mathrm{B}$
D volume of the solution is different from the sum of volume of the solute and the solvent.

The correct answer is:
attraction force remains same as in $\mathrm{A}-\mathrm{A}$ and $\mathrm{B}-\mathrm{B}$

Solution:
Raoult's law is valid for ideal solution only. These two components A and B follows the condition of Raoult's law if the force of attraction between $\mathrm{A}$ and $\mathrm{B}$ is equal to the force of attraction between A-A and B-B
Positive non-ideal behavior of the vapor pressure of a solution follows Henry's Law at low concentrations and Raoult's Law at high concentrations (pure).

Topic: Chemistry, Solutions, NEET, NEET 2002
A solution contains $\mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}$ and $\mathrm{I}^{-}$ions. This solution was treated with iodine at $35^{\circ} \mathrm{C}$. $E^{\circ}$ for $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ is $+0.77 \mathrm{~V}$ and $\mathrm{E}^{\circ}$ for $\mathrm{I}_2 / 2 \mathrm{I}^{-}=0.536 \mathrm{~V}$. The favourable redox reaction is

Options:
A $\mathrm{I}_2$ will be reduced to $\mathrm{I}^{-}$
B There will be no redox reaction
C $\mathrm{I}^{-}$will be oxidised to $\mathrm{I}_2$
D $\mathrm{Fe}^{2+}$ will be oxidised to $\mathrm{Fe}^{3+}$

The correct answer is:
$\mathrm{I}^{-}$will be oxidised to $\mathrm{I}_2$

Solution:
$2 \mathrm{I}^{-} \longrightarrow \mathrm{I}_2+2 e^{-}$(Oxidation half-reaction)
$$
E_{\text {oxi. }}^{\circ}=-0.536 \mathrm{~V} \text {. }
$$
$\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+}$ (Reduction half-reaction)
$$
\begin{aligned}
E^{\circ} & =E_{\text {oxi }}^{\circ}+E_{\text {red }}^{\circ} \\
& =+\mathrm{ve}
\end{aligned}
$$
So, reaction will take place.

Topic: Chemistry, Redox Reactions, NEET, NEET 2011 (Mains)
A solution contains non-volatile solute of molecular mass $M_2$. Which of the following can be used to calculate the molecular mass of the solute in terms of the osmotic pressure?
Note: $m_2$ is the mass of the solute, $\mathrm{V}$ is the volume of the solution and $\pi$ is the osmotic pressure.

Options:
A $\mathrm{M}_2=\left(\frac{m_2}{\pi}\right) \mathrm{VRT}$
B $\mathrm{M}_2=\left(\frac{m_2}{\mathrm{~V}}\right) \frac{\mathrm{RT}}{\pi}$
C $\mathrm{M}_2=\left(\frac{m_2}{\mathrm{~V}}\right) \pi \mathrm{RT}$
D $\mathrm{M}_2=\left(\frac{m_2}{\mathrm{~V}}\right) \frac{\pi}{\mathrm{RT}}$

The correct answer is:
$\mathrm{M}_2=\left(\frac{m_2}{\mathrm{~V}}\right) \frac{\mathrm{RT}}{\pi}$

Solution:
For dilute solution, the van't Hoff equation is:
$\begin{aligned}
\pi & =\frac{n}{\mathrm{~V}} \mathrm{RT} \\
\pi \mathrm{V} & =\frac{m_2}{\mathrm{M}} \mathrm{RT} \\
\Rightarrow \quad \mathrm{M} & =\frac{m_2 \mathrm{RT}}{\pi \mathrm{V}}
\end{aligned}$
Van't Hoff explained that when solutes are dissolved in a solvent they dissociate into ions. Since colligative properties depend only on the number of solute particles, the dissociation of solute molecules into ions results in an increase in the number of particles and hence affects the colligative properties.

Topic: Chemistry, Solutions, JEE Main
A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at $20^{\circ} \mathrm{C}$ are 440 $\mathrm{mm} \mathrm{Hg}$ for pentane and $120 \mathrm{~mm} \mathrm{Hg}$ for hexane. the mole fraction of pentane in the vapour phase would be:

Options:
A 0.200
B 0.549
C 0.786
D 0.478

The correct answer is:
0.478

Solution:
$$
\begin{aligned}
\frac{n_{\mathrm{C}_5 \mathrm{H}_{12}}}{n_{\mathrm{C}_6 \mathrm{H}_{14}}} & =\frac{1}{4} \\
\Rightarrow \mathrm{X}_{\mathrm{C}_3 \mathrm{H}_{12}} & =\frac{1}{5} \text { and } \mathrm{X}_{\mathrm{C}_6 \mathrm{H}_{14}}=\frac{4}{5} \\
\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\circ} & =440 \mathrm{~mm} \mathrm{Hg} ; \\
\mathrm{P}_{\mathrm{C}_6 \mathrm{H}_{14}}^{\circ} & =120 \mathrm{~mm} \mathrm{Hg} \\
\mathrm{P}_{\mathrm{T}} & =\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\circ} \mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}} \\
& +\mathrm{P}_{\mathrm{C}_6 \mathrm{H}_{14}}^{\circ} \mathrm{X}_{\mathrm{C}_6 \mathrm{H}_{14}} \\
& =440 \times \frac{1}{5}+120 \times \frac{4}{5} \\
& =184 \mathrm{~mm} \text { of } \mathrm{Hg}
\end{aligned}
$$
By Raoult's Law,
$$
\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}=\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\circ} \mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}}
$$
By Dalton's Law,
$$
\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}=\mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\prime} \mathrm{P}
$$
From (1) and (2),
$$
\begin{aligned}
\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}} & =440 \times \frac{1}{5}=88 \mathrm{~mm} \text { of } \mathrm{Hg} \\
\Rightarrow \quad 88 & =\mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\prime} \times 184 \\
\mathrm{X}^{\prime} & =\frac{88}{184} ; \mathrm{X}^{\prime}=0.478 .
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main
A solution is obtained by mixing 200 g of 30% and 300 g of 20% solution by weight. What is the percentage of solute in the final solution?

Options:

The correct answer is:
24

Solution:
Solute in 200 g of 30% solution =60 g
Solute in 300 g of 20% solution =60 g
total grams of solute =120 g
total grams of solution =200+300=500 g
% of solute in the final solution =Wt. of soluteWt. of solutionÃ100
% of solute in the final solution =120500Ã100=24%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A solution is obtained by mixing \( 200 \mathrm{~g} \) of \( 30 \% \) and \( 300 \mathrm{~g} \) of \( 20 \% \) solution by weight. What is the percentage of solute in the final solution?

Options:

The correct answer is:
24

Solution:
Solute in 200 g of 30% solution =60 g
Solute in 300 g of 20% solution =60 g
total grams of solute =120 g
total grams of solution =200+300=500 g
% of solute in the final solution =Wt. of soluteWt. of solution×100
% of solute in the final solution =120500×100=24%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A solution is prepared by adding 1 mole ethyl alcohol in 9 mole water. The mass percent of solute in the solution is ________ (Integer answer) (Given : Molar mass in \(\mathrm{g} \mathrm{mol}^{-1}\) Ethyl alcohol : 46 water: 18)

Options:

The correct answer is:
22

Solution:
Mass percent of Alcohol
$\begin{aligned} & =\frac{\text { Mass of ethyl alcohol }}{\text { Total mass of solution }} \times 100 \\ & =\frac{1 \times 46}{1 \times 46+9 \times 18} \times 100=\frac{4600}{208} \\ & =22.11 \text { Or } 22\end{aligned}$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (08 Apr Shift 2)
A solution is prepared by adding 2 g of ''X'' to 1 mole of water. Mass percent of ''X''in solution is

Options:
A 5%
B 20%
C 2%
D 10%

The correct answer is:
10%

Solution:
To calculate the mass percent of ''X'' in the solution, we need to determine the mass of ''X'' and the total mass of the solution.

Given:

:Solute (X) = 2 g

Solvent (H2O) = 1 mole = 18 g

Total mass = 2 + 18 = 20 g

Mass percent of X=22+18×100=10

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2023 (11 Apr Shift 2)
A solution is prepared by dissolving 0.6 g of urea (molar mass =60 g mol-1) and 1.8 g of glucose (molar mass =180 g mol-1) in 100 mL of water at 27oC. The osmotic pressure of the solution is:
(R=0.08206 L atm K-1 mol-1)

Options:
A 8.2 atm
B 2.46 atm
C 4.92 atm
D 1.64 atm

The correct answer is:
4.92 atm

Solution:
i factor of glucose and urea are 1 : 1 respectively.

Moles of glucose =weightMolecular wieght=1.8180=10-2

Moles of urea =0.660=10-2

Total mole of solute =10-2+10-2=2×10-2

Concentration of solution =2×10-2100×10-3=0.2

Osmotic pressure (π)=CRT

=0.2×0.0821×300=4.926 atm

Topic: Chemistry, Solutions, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3 . If vapour pressure of CH2Cl2 and CHCl3 at 298K are 415 and 200 mm Hg respectively, the mole fraction of CHCl3 in vapour form is: Molar mass of Cl=35.5 g mol-1

Options:
A 0.162
B 0.675
C 0.325
D 0.486

The correct answer is:
0.325

Solution:
mole of CH2Cl2 in liquid phase=8.585=0.1
mole of CHCl3 in liquid phase=11.95119.5=0.1
mole fraction of CH2Cl2 in liquid phase=0.10.2=12
mole fraction of CHCl3 in liquid phase=0.10.2=12
PT=XCH2Cl2×  vapour pressureCH2Cl2+ XCHCl3× vapour pressureCHCl3
=415× 1 2 +200× 1 2 =307.5
P T ( X A ) VP = ( X A ) LP ×  Vapour pressure of CHCl 3
307.5×  ( X CHC l 3 ) VP =200× 1 2
XCHCl3=100307.5=0.325

Topic: Chemistry, Solutions, JEE Main, JEE Main 2017 (09 Apr Online)
A solution is prepared containing a 2:1 mol ratio of dibromo ethane (C2H4Br2) and dibromo propane (C3H6Br2) what is the total vapour pressure over the solution assuming ideal behaviour?

Vapour pressure
(mm Hg)

C2H4Br2
173

C3H6Br2
127

Options:
A 300 mm Hg
B 158 mm Hg
C 150 mm Hg
D 142 mm Hg

The correct answer is:
158 mm Hg

Solution:
P=PA+PB
=PA0XA+PB0XB
=173×23+127×13
=346+1273
=4733=157.6
≈158 mm of Hg

Topic: Chemistry, Solutions, JEE Main
A solution of (-)1-chloro-1-phenylethane in toluene racemises slowly in the presence of a small amount of SbCl5, due to the formation of :

Options:
A Carbocation
B Free radical
C Carbanion
D Carbene

The correct answer is:
Carbocation

Solution:
(-) - 1 - chloro-1-phenylethane is:

.
With SbCl5 carbocation is formed. As the carbocation is planar, Cl ⊖ can again attack from both sides giving racemic mixture. (d, l mixture formed)

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2013 (07 Apr)