datavorous/entrance-exam-dataset huggingface dataset:
The values of dissociation constants of some acids (at $25^{\circ} \mathrm{C}$ ) are as follows. Indicate which is the strongest acid in water

Options:
A $1.4 \times 10^{-2}$
B $1.6 \times 10^{-4}$
C $4.4 \times 10^{-10}$
D $4.3 \times 10^{-7}$

The correct answer is:
$1.4 \times 10^{-2}$

Solution:
More is $K_a$, lesser is $p K_a\left(p K_a=-\log K_a\right)$ more is acidic strength.

Topic: Chemistry, Ionic Equilibrium, JEE Main
The values of electronegativity of atom A and B are 1.20 and 4.0 respectively. The percentage of ionic character of A-B bond is nearly _____. Roundoff answer to nearest integer.

Options:

The correct answer is:
72

Solution:
% ionic character of a bond is determined with the help of Hanny-Smith's equation.
% ionic characters =1.6( ÎEN )+3.5 ( ÎEN ) 2
Where, ÎEN = difference of electronegativity =
4.0â1.2 = 2.80
So, % ionic character =1.6( 2.80 )+3.5 ( 2.80 ) 2 =72.24%

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The values of electronegativity of atom A and B are 1.20 and 4.0 respectively. The percentage of ionic character of A-B bond is nearly

Options:

The correct answer is:
72.24

Solution:
% ionic character of a bond is determined with the help of Hanny-Smith's equation.
% ionic characters =1.6( ÎEN )+3.5 ( ÎEN ) 2
Where, ÎEN = difference of electronegativity =
4.0â1.2 = 2.80
So, % ionic character =1.6( 2.80 )+3.5 ( 2.80 ) 2 =72.24%

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The values of α, for which 132α+32113α+132α+33α+10=0, lie in the interval

Options:
A (-2,1)
B (-3,0)
C -32,32
D (0,3)

The correct answer is:
(-3,0)

Solution:
Given: 132α+32113α+132α+33α+10=0

Applying, R1→R1-R2

⇒07676113α+132α+33α+10=0

Applying, C2→C2-C3

⇒00761-αα+132α+33α+10=0

⇒0-0+763α+1+2α2+3α=0

⇒2α2+6α+1=0

⇒α=-6±36-4212×2

⇒α=-6±274

⇒α=-3±72

⇒α≈-3±2.62

⇒α≈-0.42, -5.62

⇒α≈-0.2, -2.8

⇒α∈-3,0

Topic: Mathematics, Determinants, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The values of λ for which one root of the equation x2+1-2λx+λ2-λ-2=0 is greater than 3 and the other smaller than 2 are given by

Options:
A 2<λ<5
B 1<λ<4
C 1<λ<5
D 2<λ<4

The correct answer is:
2<λ<4

Solution:
D>0, f2<0 and f3<0
D=1-2λ2-4λ2-λ-2=1+4λ2-4λ-4λ2+4λ+8
=9>0 (always true)
f2<0⇒4+21-2λ+λ2-λ-2<0
⇒4+2-4λ+λ2-λ-2<-0
⇒λ2-5λ+4<0⇒λ∈1,4 ...(i)
f3<0⇒9+31-2λ+λ2-λ-2<0
⇒9+3-6λ+λ2-λ-2<0
⇒λ2-7λ+10<0⇒λ∈2,5 ...(ii)
Taking intersection of (i) and (ii), we get,

λ∈2,4

Topic: Mathematics, Quadratic Equation, JEE Main
The values of $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ for the reaction, $\mathrm{C}$ (graphite) $+\mathrm{CO}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}(\mathrm{g})$ are $170 \mathrm{~kJ}$ and $170 \mathrm{JK}^{-1}$ respectively. This reaction will be spontaneous at -

Options:
A $510 \mathrm{~K}$
B $710 \mathrm{~K}$
C $910 \mathrm{~K}$
D $1110 \mathrm{~K}$

The correct answer is:
$1110 \mathrm{~K}$

Solution:
For a spontaneous reaction
$$
\Delta \mathrm{G}=-\mathrm{ve}
$$
Or ar eq. $\Delta \mathrm{G}=0$
$$
\begin{aligned}
\Delta \mathrm{H} & =\mathrm{T} \Delta \mathrm{S} \\
\mathrm{T} & =\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}} \\
& =\frac{170 \times 10^3}{170} \\
& =1000 \mathrm{~K}
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2009 (Mains)
The values of $K_{P 1}$ and $K_{\mathrm{P} 2}$ for the reactions
$$
\begin{aligned}
& X \rightleftharpoons Y+Z . \\
& A \rightleftharpoons 2 B . .
\end{aligned}
$$
are in ratio of 9:1. If degree of dissociation of $X$ and $\mathrm{A}$ be equal, then total pressure at equilibrium (1) and (2) are in the ratio

Options:
A $1: 1$
B $3: 1$
C $1: 9$
D $36: 1$

The correct answer is:
$36: 1$

Solution:
$$
\begin{aligned}
& \text { or } \frac{\mathrm{K}_{\mathrm{P}_1}}{\mathrm{~K}_{\mathrm{P}_2}}=\frac{\mathrm{P}}{4 \mathrm{P}^{\prime}} \quad \text { or } \quad \frac{9}{1}=\frac{\mathrm{P}}{4 \mathrm{P}^{\prime}} \\
& \text { or } \frac{\mathrm{P}^{\prime}}{\mathrm{P}}=36: 1
\end{aligned}
$$

Topic: Chemistry, Chemical Equilibrium, NEET, NEET 2008 (Mains)
The values of $m, n$, for which the system of equations
$\begin{aligned}
& x+y+z=4, \\
& 2 x+5 y+5 z=17, \\
& x+2 y+\mathrm{m} z=\mathrm{n}
\end{aligned}$
has infinitely many solutions, satisfy the equation:

Options:
A $m^2+n^2-m n=39$
B $m^2+n^2-m-n=46$
C $m^2+n^2+m+n=64$
D $m^2+n^2+m n=68$

The correct answer is:
$m^2+n^2-m n=39$

Solution:
$\begin{aligned} & \mathrm{D}=\left|\begin{array}{lll}1 & 1 & 1 \\ 2 & 5 & 5 \\ 1 & 2 & \mathrm{~m}\end{array}\right|=0 \Rightarrow \mathrm{m}=2 \\ & \mathrm{D}_3=\left|\begin{array}{ccc}1 & 1 & 4 \\ 2 & 5 & 17 \\ 1 & 2 & \mathrm{n}\end{array}\right|=0 \Rightarrow \mathrm{n}=7\end{aligned}$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The values of $p$ and $q$ so that the function
$$
f(x)=\left\{\begin{array}{cl}
(1+\mid \sin x)^{\frac{p}{\sin x}}, \frac{-\pi}{6} < x < 0 \\
q & , x=0 \\
e^{\frac{\sin 2 x}{\sin 3 x}} & , 0 < x < \frac{\pi}{6}
\end{array}\right.
$$
is continuous at $x=0$, are

Options:
A $p=\frac{1}{3}, q=e^{2 / 3}$
B $p=0, q=e^{2 / 3}$
C $p=\frac{2}{3}, q=e^{-2 / 3}$
D $p=-\frac{2}{3}, q=e^{2 / 3}$

The correct answer is:
$p=-\frac{2}{3}, q=e^{2 / 3}$

Solution:
$$
\begin{aligned}
\text { At } x & =0, \text { LHL }=\lim _{x \rightarrow 0}(1+|\sin x|)^{\frac{\rho}{\sin x}} \\
& =\lim _{x \rightarrow 0}(1-\sin x)^{\frac{\rho}{\sin x}} \\
& \quad\left[\because|\sin x|=-\sin x, \text { for } \frac{-\pi}{6} < x < 0\right] \\
& =\lim _{h \rightarrow 0}(1+\sin h)^{\frac{-\rho}{\sin h}} \\
& =e^{\lim _{h \rightarrow 0}(\sin h)\left(\frac{-\rho}{\sin h}\right)}=e^{-\rho} \\
\text { RHL } & =\lim _{x \rightarrow 0^{+}} e^{\frac{\sin 2 x}{\sin 3 x}}
\end{aligned}
$$
$$
\begin{aligned}
= & e^{x \rightarrow 0^{+} \frac{\sin 2 x}{\sin 3 x}}=e^{\lim _{h \rightarrow 0} \frac{\sin 2 h}{\sin 3 h}} \\
= & e^{\lim _{h \rightarrow 0} \frac{\sin 2 h}{2 h} \cdot \frac{1}{\frac{\sin 2 h}{3 h}} \cdot \frac{2}{3}}=e^{2 / 3} \\
f(0)= &
\end{aligned}
$$
Since, $f(x)$ is continuous at $x=0$.
$$
\begin{array}{ll}
\therefore & \mathrm{LHL}=\mathrm{RHL}=f(0) \\
\Rightarrow & \mathrm{e}^{-P}=\mathrm{e}^{2 / 3}=q \\
\Rightarrow & p=\frac{-2}{3}, q=e^{2 / 3}
\end{array}
$$

Topic: Mathematics, Continuity and Differentiability, JEE Main
The values of the crystal field stabilization energies for a high spin d6 metal ion in octahedral and tetrahedral fields, respectively, are:

Options:
A −0.4Δo and −0.6Δt
B −2.4Δo and −0.6 Δt
C −1.6Δo and −0.4Δt
D −0.4Δo and −0.27Δt

The correct answer is:
−0.4Δo and −0.6Δt

Solution:
For d6 configuration, high spin

(i) In case of octahedral complex tgg2.1.1,eg1.1

CFSE=-0.4nt2g+0.6negΔ0+n(P)

=[-0.4×4+0.6×2]Δ0+0

=-0.4Δ0

(i) In case of tetrahedral complex e2,1,t21,1,1

CFSE=-0.6ne+0.4nt2Δr

=[-0.6×3+0.4×3]Δt

=-0.6Δt

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (05 Sep Shift 1)
The values of two resistors are R1=6 Â±0.3 kÎ© and R2=10 Â±0.2 kÎ©. The percentage error in the equivalent resistance when they are connected in parallel is

Options:
A 2%
B 3.125%
C 7%
D 10.125%

The correct answer is:
10.125%

Solution:
R1=6Â±0.3kÎ©, R2=10Â±0.2kÎ©
Rparallel=R1R2R1+R2
Let R1+R2=x
â RP=R1R2x
Taking log of both sides
ln RP=ln R1+lnâ¡R2-lnâ¡x
Differentiating,
âRPRP=âR1R1+âR2R2+-âxx
âxmean=0.3+0.22=0.25 Î©
Rmean=6+102=8 Î©
â´ x=6+102=8 Î©
â âxx=0.258
â´Total error=0.36+0.210+0.258
=0.05+0.02+0.03125=0.10125
â´ âRPRP=10.125%

Topic: Physics, Units and Dimensions, JEE Main
The values of $x$ and $y$ for which the numbers $3+i x^2 y$ and $x^2+y+4 i$ are conjugate complex can be

Options:
A $(-2,-1)$ or $(2,-1)$
B $(-1,2)$ or $(-2,1)$
C $(1,2)$ or $(-1,-2)$
D None of these

The correct answer is:
$(-2,-1)$ or $(2,-1)$

Solution:
According to condition, $\quad 3-i x^2 y=x^2+y+4 i$
$\begin{aligned}& \Rightarrow x^2+y=3 \text { and } x^2 y=-4 \Rightarrow x= \pm 2, y=-1 \\& \Rightarrow(x, y)=(2,-1) \text { or }(-2,-1)\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
The values of $x$ and $y$ satisfying the equation $\frac{(1+i) x-2 i}{3+i}+\frac{(2-3 i) y+i}{3-i}=i$ are

Options:
A $x=-1, y=3$
B $x=3, y=-1$
C $x=0, y=1$
D $x=1, y=0$

The correct answer is:
$x=3, y=-1$

Solution:
$\begin{aligned} & \frac{(1+i) x-2 i}{3+i}+\frac{(2-3 i) y+i}{3-i}=i \\ & \quad \Rightarrow(4+2 i) x+(9-7 i) y-3 i-3=10 i\end{aligned}$
Equating real and imaginary parts, we get $2 x-7 y=13$ and $4 x+9 y=3$. Hence $x=3$ and $y=-1$.
Trick : After finding the equations, no need to solve them, put the values of $x$ and ${ }^y$ given in the options and get the appropriate option.

Topic: Mathematics, Complex Number, JEE Main
The value(s) of x satisfying the equation sin-1â¡1-x-2sin-1â¡x=Ï2 is/are

Options:
A 0
B 12
C 0,12
D -12

The correct answer is:
0

Solution:
We have,
sin-1⁡1-x-2sin-1⁡x=π2
⇒sin-1⁡1-x=π2+2sin-1⁡x
⇒1-x=sin⁡π2+2sin-1⁡x
⇒1-x=cos⁡2sin-1⁡x
⇒1-x=cos⁡cos-1⁡1-2x2∵2sin-1⁡x=cos-1⁡1-2x2
⇒1-x=1-2x2
⇒x=2x2⇒x2x-1=0⇒x=0,12
For, x=12, we have
LHS =sin-1⁡1-x-2sin-1⁡x
=sin-1⁡12-2sin-1⁡12=-sin-1⁡12=-π6≠R.H.S.
So, x=12 is not a root of the given equation.
Clearly, x=0 satisfies the equation
Here, x=0 is the root of the given equation.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The van der Waal's constant 'b' of a gas is 4 π × 1 0 - 4 L/mol . How near can the centres of the two molecules approach each other ? [Use : NA=6×1023]

Options:
A 1 0 - 7 m
B 1 0 - 10 m
C 5 × 1 0 - 1 1 m
D 5 × 1 0 - 9 m

The correct answer is:
1 0 - 10 m

Solution:
b = 4 Ã 4 3 Ã Ï r 3 Ã N A ;
We know that, 1L=1000 cm3
4 Ã Ï Ã 1 0 - 4 Ã 1 0 0 0 = 4 Ã 4 3 Ã Ï Ã r 3 Ã 6 Ã 1 0 2 3
r = 5 Ã 1 0 - 9 cm
Distance of closest approach =2r
= 1 0 - 8 cm or 1 0 - 10 m

Topic: Chemistry, States of Matter, JEE Main
The van der Waals' equation for a gas is
$$
\left(P+\frac{a}{V^2}\right)(V-b)=n R T
$$
where $P, V, R, T$ and $n$ represent the pressure, volume, universal gas constant, absolute temperature and number of moles of a gas, respectively. $a$ and $b$ are constants.
The ratio $\frac{b}{a}$ will have the following dimensional formula.

Options:
A $\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2\right]$
B $\left[\mathrm{M}^{-1} \mathrm{~L}^{-1} \mathrm{~T}^{-1}\right]$
C $\left[\mathrm{ML}^2 \mathrm{~T}^2\right]$
D $\left[\mathrm{MLT}^{-2}\right]$

The correct answer is:
$\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2\right]$

Solution:
van der Waals gas equation is
$$
\left(p+\frac{a}{V^2}\right)(V-b)=n R T
$$
From principle of homogeneity
Dimension of $\frac{a}{V^2}=$ dimension of $P$
$\therefore$ Dimension of $a=\left[V^2\right] \times[P]$
$$
\begin{aligned}
{[a] } & =\left[\mathrm{L}^3\right]^2\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \\
& =\left[\mathrm{ML}^5 \mathrm{~T}^{-2}\right]
\end{aligned}
$$
Dimension of $b=$ dimension of $V$
$$
\begin{aligned}
& {[b]=\left[\mathrm{L}^3\right]} \\
& \therefore \quad \frac{b}{a}=\frac{L^3}{\mathrm{ML}^5 \mathrm{~T}^{-2}}=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2\right] \\
&
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main
The van der Waals equation for one mole of a real gas can be written as P + aV2V â b=RT . For the gases H2 , NH3 , and CH4 , the value of 'a' bar Lâ2molâ2 are 0.2453, 4.170 and 2.253 respectively.
Which of the following can be inferred from the 'a' values?

Options:
A NH3 can be most easily liquified
B H2 can be most easily liquified
C value of 'a' for CH4 is less than that of NH3 because it has the lower molar mass
D intermolecular forces are the strongest in hydrogen

The correct answer is:
NH3 can be most easily liquified

Solution:
Van der Wall's constant 'a' gives information about force of attraction between gaseous molecules. Higher the value of 'a' higher will be force of attraction, and easier the liquefaction of gas. Order of intermolecular force of attraction = H2< CH4< NH3

Topic: Chemistry, States of Matter, JEE Main
The Vander Waal's constant ' $a$ ' for the gases $\mathrm{O}_2, \mathrm{~N}_2, \mathrm{NH}_3$ and $\mathrm{CH}_4$ are $1.3,1.390,4.170$ and $2.253 \mathrm{~L}^2$ atmmol $^{-2}$ respectively. The gas which can be most easily liquefied is

Options:
A $\mathrm{O}_2$
B $N_2$
C $\mathrm{NH}_3$
D $\mathrm{CH}_4$

The correct answer is:
$\mathrm{NH}_3$

Solution:
Value of constant $a$ is greater than other for $\mathrm{NH}_3$ that's why $\mathrm{NH}_3$ can be most easily liquefied.

Topic: Chemistry, States of Matter, JEE Main
The Van't Hoff factor will be highest for

Options:
A Sodium chloride
B Magnesium chloride
C Sodium phosphate
D Urea

The correct answer is:
Sodium phosphate

Solution:
$\mathrm{Na}_3 \mathrm{PO}_4$ gives maximum four ion it is show highest Vant's haff factor.

Topic: Chemistry, Solutions, JEE Main
The vapor pressure of benzene is 53.3 kPa at 603 oC, but it falls to 51.5  kP a when 19 g of a non-volatile organic compound is dissolved in 500 g benzene. The molar mass of the non-volatile compound is

Options:
A 82
B 85
C 88
D 92

The correct answer is:
85

Solution:
P 0  = 53.3 kPa
P s =51.5  kP a
P0-PsPs=nN=19Ã78MÃ500
53.3-51.551.5=nN=19Ã78MÃ500
M=19Ã78Ã51.5(53.3-51.5)Ã500
M=85 g

Topic: Chemistry, Solutions, JEE Main
The vapour density of N2O4 at a certain temperature is 30. What is the percentage dissociation of N2O4 at this temperature ?

Options:
A 53.3 %
B 106.6 %
C 26.7 %
D None of these

The correct answer is:
53.3 %

Solution:
The reaction is
N 2 O 4 â 2 NO 2
mol. wt. of N2O4 = 14 x 2 + 16 x 4 = 92
â´ Vapour density (D) of N2O4 = 92/2 = 46
We know that,
Î± = D - d n - 1 d = 4 6 - 3 0 2 - 1 3 0 = 1 6 3 0
where,
D = Initial vapour density of reactant
d = vapour density of equilibrium mixture
α= degree of dissociation
â´ Î± = 0.533
= 53.3 %

Topic: Chemistry, Chemical Equilibrium, JEE Main
The vapour of Hg absorb some electrons accelerated by a potential difference of 4.5 volts from rest as a result of which light is emitted. If the full energy of a single incident electron is supposed to be converted into light emitted by electron in a single Hg atom, find the wave number 1λ of the light

Options:
A 3.63Ã106 m-1
B 5.93Ã106Â m-1
C 5.93Ã106Â cm-1
D 5.62Ã106Â m-1

The correct answer is:
3.63Ã106 m-1

Solution:
Potential difference V=4.5 eV
â´ Potential energy =4.5 eV energy absorbed and this energy is emitted by electron of Hg atom.
â´ E=hcÎ» =4.5 eV
=4.5Ã1.6Ã10-19 J
â´ 1Î»=4.5Ã1.6Ã10-196.6Ã10-34Ã3.8Ã108 m-1 â3.63Ã106 m-1

Topic: Chemistry, Structure of Atom, JEE Main
The vapour pressure of a non-ideal tw component solution is given below
Identify the correct $T-X$ curve for the same mixture,

Options:
A

B

C

D

The correct answer is:

Solution:
No solution. Refer to answer key.

Topic: Chemistry, Solutions, JEE Main
The vapour pressure of a pure liquid A is 70 torr at 27°C . It forms an ideal solution with another liquid B. The mole fraction of B is 0.2 and total vapour pressure of the solution in 84 torr at 27°C . The vapour pressure of pure liquid B at 27°C is

Options:
A 140 torr
B 56 torr
C 14 torr
D 70 torr

The correct answer is:
140 torr

Solution:
According to Raoult's law, p=pAo.XA+pBo.XB
Here pAo=70 torr,XA=1-XB=1-0.2=0.8,p=84 torr, XB=0.2 and pBo=?
So 84 torr =70 torr×0.8+pBo×0.2
or pBo=84 torr-56 torr0.2=140 torr .

Topic: Chemistry, Solutions, JEE Main
The vapour pressure of acetone at 20oC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20oC , its vapour pressure was 183 torr. The molar mass (g mol-1) of the substance is:

Options:
A 488
B 32
C 64
D 128

The correct answer is:
64

Solution:
Δ P = 1 8 5 - 1 8 3 = 2 torr
MCH3-CO||-CH3=15×2+16+12=58 g/mol
ΔPP0=185-183185=2185=XB=1.2M1.2M+10058
1.2 M << 1 0 0 5 8
⇒ 2 1 8 5 = 1.2 M × 5 8 1 0 0
M = 5 8 × 1.2 1 0 0 × 1 8 5 2
=64.38≈64 g/mol

Topic: Chemistry, Solutions, JEE Main, JEE Main 2015 (04 Apr)
The vapour pressure of benzene at a certain temperature is $640 \mathrm{~mm}$ of $\mathrm{Hg}$. A non-volatile and non-electrolyte solid weighing $2.175 \mathrm{~g}$ is added to $39.08 \mathrm{~g}$ of benzene. The vapour pressure of the solution is $600 \mathrm{~mm}$ of $\mathrm{Hg}$. What is the molecular weight of solid substance

Options:
A 49.50
B 59.6
C 69.5
D 79.8

The correct answer is:
69.5

Solution:
\(\begin{aligned}
& \frac{p^{\circ}-p}{p^\circ}=x_B=\frac{n_B}{n_A} \\
& \frac{640-600}{640}=\frac{2.175 \times 78}{M_B \times 39.08} \\
& M_B=\frac{2.175 \times 78 \times 640}{39.08 \times 40} \\
& =69.5
\end{aligned}\)

Topic: Chemistry, Solutions, JEE Main
The vapour pressure of benzene at a certain temperature is $640 \mathrm{~mm}$ of $\mathrm{Hg} .$ A non-volatile and non-electrolyte solid weighing $2.175 \mathrm{~g}$ is added to $39.08 \mathrm{~g}$ of benzene. If the vapour pressure of the solution is $600 \mathrm{~mm}$ of $\mathrm{Hg}$, what is the molecular weight of solid substance?

Options:
A $49.50$
B $59.60$
C $69.60$
D $79.82$

The correct answer is:
$69.60$

Solution:
Given, vapour pressure of benzene,
$$
p^{\circ}=640 \mathrm{~mm} \mathrm{Hg}
$$
Vapour pressure of solution,
$$
p=600 \mathrm{~mm} \mathrm{Hg}
$$
Weight of solute, $w=2.175 \mathrm{~g}$ Weight of benzene, $W=39.08 \mathrm{~g}$ Molecular weight of benzene,
$$
M=78 \mathrm{~g}
$$
Molecular weight of solute, $m=?$ According to Raoult's law,
$\begin{aligned} \frac{p^{\circ}-p}{p^{\circ}} &=\frac{w \times M}{m \times W} \\ \frac{640-600}{640} &=\frac{2.175 \times 78}{m \times 39.08} \\ \frac{40}{640} &=\frac{2.175 \times 78}{m \times 39.08} \\ m &=\frac{16 \times 2.175 \times 78}{39.08} \\ m &=69.60 \end{aligned}$

Topic: Chemistry, Solutions, JEE Main
The vapour pressure of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of toluene in vapour phase in cotanct with equimolar solution of benzene and toluene is

Options:
A 0.50
B 0.6
C 0.27
D 0.73

The correct answer is:
0.27

Solution:
In equimolar solution of benzene and toluene, mole fraction of benzene,
Xb= mole-fraction of toluene, Xt=11+1=0.5
So partial pressure of benzene,
pb=Xb.pbo=0.5×160 torr = 80 torr and partial pressure of toluene,
pt=Xtpto=0.5×60 torr = 30 torr
Hence, vapour pressure of solution = 80 torr + 30 torr = 110 torr
According to Dalton's law of partial pressure,
Partial vapour pressure of toluene,
pt= mole fraction of touene in vapour phase × total pressure ptotal
or mole fraction of toluene in vapour phase =ptptotal=30 torr110 torr=0.27

Topic: Chemistry, Solutions, JEE Main
The vapour pressure of pure liquid A is 10 torr and at the same temperature when 9 g of B solid dissolved in 20 g of A, its Vapour pressure is reduced to 5 torr. If the molecular mass of A is 200 amu, then the molecular mass of B is

Options:
A 100 amu
B 90 amu
C 75 amu
D 120 amu

The correct answer is:
90 amu

Solution:
P0-PsPs=nN
10-55=9Ã200MÃ20
M = 90 amu

Topic: Chemistry, Solutions, JEE Main
The vapour pressure of pure water at 26oC is 25.5 torr. . The vapour pressure of a solution which contains 20.0 glucose, (C 6 H 12 O 6 ) , in 100 g water (in torr) is?

Options:

The correct answer is:
25

Solution:
By raoult's law
PAÎ¿-PAPAÎ¿=nBnA+nB
Or PAÎ¿-PAPA=nBnA
PA := the partial vapour pressure of the component {display style i} in the gaseous mixture (above the solution),
PAÎ¿ = the vapour pressure of the pure component
nB : moles of solute
nA : moles of solvent
PAÎ¿-PAPA=nBnA
25.5-PAPA=2018010018
25.5-PAPA=20Ã18100Ã180=0.02
25.5− P A =0.02  P A
1.02  P A =25.5
PA=25.0

Topic: Chemistry, Solutions, JEE Main
The vapour pressure of two liquids $P$ and $\mathrm{Q}$ are 80 and 60 torr respectively. The total vapour pressure of solution obtained by mixing 3 mole of $\mathrm{P}$ and $2 \mathrm{~mol}$ of $\mathrm{Q}$ would be:

Options:
A 72 torr
B 140 torr
C 68 torr
D 20 torr

The correct answer is:
72 torr

Solution:
By Raoult's Law $\mathrm{P}_{\mathrm{T}}=\mathrm{P}_{\mathrm{P}}^{\circ} \mathrm{X}_{\mathrm{P}}+\mathrm{P}_{\mathrm{Q}}^{\circ} \mathrm{X}_{\mathrm{Q}}$
$$
\begin{aligned}
& \mathrm{P}_{\mathrm{P}}^{\circ}=80 \text { torr } \\
& \mathrm{X}_{\mathrm{P}}=\frac{3}{5} \\
& \mathrm{P}_{\mathrm{Q}}^{\circ}=60 \text { torr }
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{X}_{\mathrm{Q}} & =\frac{2}{5} \\
\mathrm{P}_{\mathrm{T}} & =80 \times \frac{3}{5}+60 \times \frac{2}{5} \\
& =48+24 \\
\mathrm{P}_{\mathrm{T}} & =72 \text { torr. }
\end{aligned}
$$
Related Theory
Solutes which dissociate or associate in a particular solution will not obey Raoult's law. Raoult's law applies to solutions containing non-volatile solute only. Raoult's law cannot be applied to concentrated solutions.

Topic: Chemistry, Solutions, JEE Main
The vapour pressure of two liquids $P$ and $Q$ are 80 and 600 torr, respectively. The total vapour pressure of solution obtained by mixing 3 mole of $P$ and 2 mole of $Q$ would be

Options:
A 140 torr
B 20 torr
C 68 torr
D 72 torr

The correct answer is:
72 torr

Solution:
$\begin{aligned} & P_T=P_P^0 X_P+P_Q^0 X_Q ; \quad P_T=80 \times \frac{3}{5}+60 \times \frac{2}{5} \\ & P_T=48+24=72 \text { torr } .\end{aligned}$

Topic: Chemistry, Solutions, JEE Main
The vapour pressure of two pure liquids $A$ and $B$ that form an ideal solution, are 400 and $800 \mathrm{~mm}$ of Hg respectivelyat a temperature $t^{\circ} \mathrm{C}$. The mole fraction of $A$ in a solution of $A$ and $B$ whose boiling point is $t^{\circ} \mathrm{C}$ will be

Options:
A 0.4
B 0.8
C 0.1
D 0.2

The correct answer is:
0.1

Solution:
V.P. of solution at $t^{\circ} \mathrm{C}=760 \mathrm{~mm}$ [at b.p., V.P. of solution =atompheric pressure] Thus $=\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{x}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^{\infty} \mathrm{x}_{\mathrm{B}}$

or $\mathrm{P}=\mathrm{P}_{\mathrm{A}}^{0} \cdot \mathrm{x}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^{0}\left(1-\mathrm{x}_{\mathrm{A}}\right)\left[\because \mathrm{x}_{\mathrm{A}}+\mathrm{x}_{\mathrm{B}}=1\right]$

or $760=400 \mathrm{X}_{\mathrm{A}}+800\left(1-\mathrm{X}_{\mathrm{A}}\right)[\because \mathrm{P}=760 \mathrm{mmofHg}]$

or $-800+760=-400 x_{\mathrm{A}}$

or $-40=-400 \mathrm{x}_{\mathrm{A}}$

or $x_{\mathrm{A}}=\frac{40}{400}=0.1$

Thus mole fraction in solution is 0.1

Topic: Chemistry, Solutions, JEE Main
The vapour pressure of water at $20^{\circ} \mathrm{C}$ is $17.5 \mathrm{~mm} \mathrm{Hg}$. If $18 \mathrm{~g}$ of glucose $\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)$ is added to $178.2 \mathrm{~g}$ of water at $20^{\circ} \mathrm{C}$, the vapour pressure of the resulting solution will be

Options:
A
$17.675 \mathrm{~mm} \mathrm{Hg}$

B
$15.750 \mathrm{~mm} \mathrm{Hg}$

C
$16.500 \mathrm{~mm} \mathrm{Hg}$

D
$17.325 \mathrm{~mm} \mathrm{Hg}$

The correct answer is:
$17.325 \mathrm{~mm} \mathrm{Hg}$

Solution:
$$
\begin{aligned}
& \frac{P^0-P_s}{P_s}=X_{\text {solute }} \\
& \frac{17.5-P_s}{P_s}=\frac{0.1}{10} \\
& \frac{17.5-P_s}{P_s}=0.01 \\
& \Rightarrow P_s=17.325 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2008
The vapour pressure of water at $23^{\circ} \mathrm{C}$ is $19.8 \mathrm{~mm}, 0.1$ mole of glucose is dissolved in $178.2 \mathrm{~g}$ of water. What is the vapour pressure (in $\mathrm{mm}$ ) of the resultant solution?

Options:
A 19.0
B 19.602
C 19.402
D 19.202

The correct answer is:
19.602

Solution:
Given $P_s=19.8 \mathrm{~mm}$
$\begin{aligned}
& n_A=0.1 \\
& n_B=\frac{178.2}{18}=9.9
\end{aligned}$
According to Raoult's law
$\begin{gathered}
\frac{P_s-P}{P_s}=\frac{n_A}{n_A+n_B} \\
\frac{19.8-P}{19.8}=\frac{0.1}{9.9+0.1}
\end{gathered}$
or
$\begin{aligned}
198-10 P & =19.8 \times 0.1 \\
10 P & =198-1.98 \\
10 P & =196.02 \\
P & =19.602 \mathrm{~mm}
\end{aligned}$

Topic: Chemistry, Solutions, JEE Main