datavorous/entrance-exam-dataset huggingface dataset:
10 mL of 1 mM surfactant solution forms a monolayer covering 0.24 cm2 on a polar substrate. If the polar head is approximated as a cube, what is its edge length?

Options:
A 1.0 pm
B 0.1 nm
C 2.0 pm
D 2.0 nm

The correct answer is:
2.0 pm

Solution:
Millimoles of surfactants =10×10-3=10-2
Moles of surfactants =10-5
Number of molecules =6×1023×10-5=6×10+18
Surface area occupied by one molecule =0.246×1018=0.04×10-18cm2
4×10-20=a2
a=2×10-10cm=2pm

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2019 (09 Apr Shift 2)
\( 1.0 \mathrm{~g} \) of \( \mathrm{Mg} \) is burnt with \( 0.28 \mathrm{~g} \) of \( \mathrm{O}_{2} \) in a closed vessel. Which reactant is left in excess and
how much?

Options:
A \( \mathrm{Mg}, 5.8 \mathrm{~g} \)
B Mg, \( 0.58 \mathrm{~g} \)
C \(\mathrm{O}_{2}, 0.24 \mathrm{~g} \)
D \( \mathrm{O}_{2}, 2.4 \mathrm{~g} \)

The correct answer is:
Mg, \( 0.58 \mathrm{~g} \)

Solution:
For the reaction
$2 \mathrm{Mg}+\mathrm{O}_{2} \rightarrow 2 \mathrm{M} \mathrm{gO}$
$48 \mathrm{~g}$ of $\mathrm{Mg}$ requires $32 \mathrm{~g}$ of $\mathrm{O}_{2}$
Therefore, $1 \mathrm{~g}$ of $\mathrm{Mg}$ require $\frac{32}{48} \mathrm{~g}$ of $\mathrm{O}_{2}=0.66 \mathrm{~g}$
But available $\mathrm{O}_{2}$ is $0.28 \mathrm{~g}$
So, $\mathrm{O}_{2}$ is the limiting reagent and $\mathrm{Mg}$ is the excess reagent.
$32 \mathrm{~g}$ of $\mathrm{O}_{2}$ needs $48 \mathrm{~g}$ of $\mathrm{Mg}$
$1 \mathrm{~g}$ of $\mathrm{O}_{2}$ needs $\frac{48}{32} \mathrm{~g}$ of $\mathrm{Mg}$
$0.28$ of $\mathrm{O}_{2}$ needs $\frac{48}{32} \times 0.28=0.42 \mathrm{~g}$
Mg available $=1 \mathrm{~g}$
Mg left $=1-0.42=0.58 \mathrm{~g}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
\( 10 \mathrm{~mL} \) of gaseous hydrocarbon on combustion, gives \( 40 \mathrm{~mL} \) of \( \mathrm{CO}_{2} \) ( \( \mathrm{g} \) ) and \( 50 \mathrm{~mL} \) of \( \mathrm{H}_{2} \mathrm{O} \) (vapour). The hydrocarbon
is:

Options:
A \( \mathrm{C}_{4} \mathrm{H}_{5} \)
B \( \mathrm{C}_{8} \mathrm{H}_{10} \)
C \( \mathrm{C}_{4} \mathrm{H}_{8} \)
D \( \mathrm{C}_{4} \mathrm{H}_{10} \)

The correct answer is:
\( \mathrm{C}_{4} \mathrm{H}_{10} \)

Solution:
The general equation for burning of hydrocarbon is given as CxHy.

CxHy+O2excess→xCO2+y2H2O.

According to Avogadro's hypothesis, gases react with each other in the same proportion of volume as moles. From this, we can write the equation as:

10CxHy+O2excess→40CO2+50H2O 1CxHy+O2excess→4CO2+5H2O.

Comparing with the general equation, we get:

x=4 and y2=5.

So, y=10.

Hence, the hydrocarbon is C4H10.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
10 mL citric acid (H3C6H5O7) is neutralised completely by 35.6 mL of 0.312 M NaOH solution. The molarity of the solution of citric acid is

Options:
A 1.11 M
B 0.45 M
C 0.11 M
D 0.37 M

The correct answer is:
0.37 M

Solution:
H3C6H5O7+3NaOHâNa3C6H5O7+3H2O
M1V11=M2V23 or M1=M2V23V1
= 0.312×35.6 3×10 =0.37 M

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
10 mL of 1 mM surfactant solution forms a monolayer covering 0.24 cm2 on a polar substrate. If the polar head is approximated as cube, what is its edge length?

Options:
A 2.0 pm
B 2.0 nm
C 1.0 pm
D 0.1 nm

The correct answer is:
2.0 pm

Solution:
Millimoles =10×10-3=10-2
Moles =10-5
No. of molecules =6×1023×10-5=6×1018
Surface area ocupied by one molecule
=0.246×1018=0.04×10-18cm2
4×10-20=x2[∵x= edge length ]
x=2×10-10 cm =2 pm

Topic: Chemistry, Surface Chemistry, JEE Main
$10 \mathrm{~mL}$ of $2(\mathrm{M}) \mathrm{~NaOH}$ solution is added to $200 \mathrm{~mL}$ of $0.5$ $\mathrm{(M)}$ of $\mathrm{NaOH}$ solution. What is the final concentration ?

Options:
A
$0.57(\mathrm{M})$

B
$5.7(\mathrm{M})$

C
$11.4(\mathrm{M})$

D
$1.14(\mathrm{M})$

The correct answer is:
$0.57(\mathrm{M})$

Solution:
From molarity equation
$\begin{aligned}
& \mathrm{M}_1 \mathrm{V}_1+\mathrm{M}_2 \mathrm{V}_2=\mathrm{MV}_{\text {(total) }} \\
& 2 \times \frac{10}{1000}+0.5 \times \frac{200}{1000}=\mathrm{M} \times \frac{210}{1000} \\
& 120=\mathrm{M} \times 210 \\
& \mathrm{M}=\frac{120}{210}=0.57 \mathrm{~M}
\end{aligned}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2013 (25 Apr Online)
10 ml of a gaseous hydrocarbon was burnt completely in 80 ml of O2 at NTP. The remaining gas occupied 70 ml at NTP. This volume become 50 ml on treatment with KOH solution. The molecular formula of the hydrocarbon is

Options:
A C2H4
B CH4
C C2H2
D C2H6

The correct answer is:
C2H4

Solution:
Suppose the molecular formula of hydrocarbon = CxHy
C x H y g + x+ y 4 O 2 g → xCO 2 g + y 2 H 2 O l
Volume of hydrocarbon = 10 ml
Decreases in volume = 10 + 80 – 70 = 20 ml
Volume of CO2=70-50=20 ml
x=2010=2
V O 2 left =50 ml
V O 2 Consumed =80−50= 30 ml
V O 2 Consumed = V H.C. × x+ y 4
30=10× 2+ y 4
y= 3−2 ×4=4
Molecular formula is C2H4 .

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
10 ml of a gaseous hydrocarbon was burnt completely in 80 ml of O2 at NTP. The remaining gas occupied 70 ml at NTP. This volume become 50 ml on treatment with KOH solution. The molecular formula of the hydrocarbon is

Options:
A C2H4
B CH4
C C2H2
D C2H6

The correct answer is:
C2H4

Solution:
Suppose the molecular formula of hydrocarbon = CxHy
C x H y g + x+ y 4 O 2 g → xCO 2 g + y 2 H 2 O l
Volume of hydrocarbon = 10 ml
Decreases in volume = 10 + 80 – 70 = 20 ml
Volume of CO2=70-50=20 ml
x=2010=2
V O 2 left =50 ml
V O 2 Consumed =80−50= 30 ml
V O 2 Consumed = V H.C. × x+ y 4
30=10× 2+ y 4
y= 3−2 ×4=4
Molecular formula is C2H4 .

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
10 moles of A2 , 15 moles of B2 and 5 moles of AB are placed in a 2 litre vessel and allowed the come to equilibrium. The final concentration of AB is 10.5 M ,

A 2 g + B 2 g â 2AB g

Determine the value of equilibrium constant KC for the reaction.

Options:
A 25.3
B 31.5
C 36.3
D 40.5

The correct answer is:
31.5

Solution:
5+2x2=10.5 ;x=8
KC=[AB]2A2B2=5+2x2210-x215-x2=(21)22×7=31.5

Topic: Chemistry, Chemical Equilibrium, JEE Main
10 moles of a mixture of gydogen and oxygen gases at a pressure of $1 \mathrm{~atm}$ at constant volume and temperature, react to form $3.6 \mathrm{~g}$ of liquid water. The pressure of the resulting mixture will be closest to

Options:
A $1.07 \mathrm{~atm}$
B $0.97$ atm
C $1.02 \mathrm{~atm}$
D $0.92 \mathrm{~atm}$

The correct answer is:
$0.97$ atm

Solution:
$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(I)$
$0.2$ mole $0.1$ mole $0.2$ mole
moles of gas remaining $=9.7$ at constant $(T) \&(V)$
$\begin{array}{l}
\frac{n_{1}}{n_{2}}=\frac{p_{1}}{p_{2}} \\
\frac{10}{9.7}=\frac{1}{p_{2}} \& p_{2}=0.97
\end{array}$

Topic: Chemistry, Surface Chemistry, KVPY, KVPY 2014 (SA)
10 resistors each of resistance 10 Ω can be connected in such as to get maximum and minimum equivalent resistance. The ratio of maximum and minimum equivalent resistance will be________.

Options:

The correct answer is:
100

Solution:
When all the resistors are connected in series, the total resistance is

Rmax=10R

⇒Rmax=10×10=100 Ω

When all the resistors are connected in parallel, the total resistance is

Rmin=R10

⇒Rmin=1010=1 Ω

Hence, the ratio,

RmaxRmin=100 Ω1 Ω=100.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (10 Apr Shift 1)
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots \ldots \ldots+\ldots \ldots \cdot \frac{1}{n \cdot(n+1)}$

Options:
A $\frac{1}{n(n+1)}$
B $\frac{n}{n+1}$
C $\frac{2 n}{n+1}$
D $\frac{2}{n(n+1)}$

The correct answer is:
$\frac{n}{n+1}$

Solution:
$\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots \ldots \ldots+\left(\frac{1}{n}-\frac{1}{n+1}\right)$
$=1-\frac{1}{n+1}=\frac{n}{n+1}$

Topic: Mathematics, Sequences and Series, JEE Main
[12]
A solid ball of radius $\mathrm{R}$ has a charge density $\rho$ given by $\rho=\rho_0\left(1-\frac{r}{R}\right)$ for $0 \leq r \leq R$. The electric field outside the ball is:

Options:
A
$\frac{\rho_0 R^3}{\varepsilon_0 r^2}$

B
$\frac{4 \rho_0 R^3}{3 \varepsilon_0 r^2}$

C
$\frac{3 \rho_0 R^3}{4 \varepsilon_0 r^2}$

D
$\frac{\rho_0 R^3}{12 \varepsilon_0 r^2}$

The correct answer is:
$\frac{\rho_0 R^3}{12 \varepsilon_0 r^2}$

Solution:
Charge density, $\rho=\rho_0\left(1-\frac{r}{R}\right)$
$$
\begin{aligned}
&d q=\rho d v \\
&q_{\text {in }}=\int d q=\rho d v \\
&=\rho_0\left(1-\frac{r}{R}\right) 4 \pi r^2 d r \quad\left(\because d v=4 \pi \mathrm{r}^2 \mathrm{dr}\right) \\
&=4 \pi \rho_0 \int_0^R\left(1-\frac{r}{R}\right) r^2 d r \\
&=4 \pi \rho_0 \int_0^R r^2 d r-\frac{r^2}{R} d r \\
&=4 \pi \rho_0\left[\left[\frac{r^3}{3}\right]_0^R-\left[\frac{r^4}{4 R}\right]_0^R\right] \\
&=4 \pi \rho_0\left[\frac{R^3}{3}-\frac{R^4}{4 R}\right]
\end{aligned}
$$
$=4 \pi \rho_0\left[\frac{R^3}{3}-\frac{R^3}{4}\right]=4 \pi \rho_0\left[\frac{R^3}{12}\right]$
$q=\frac{\pi \rho_0 R^3}{3}$
$$
E .4 \pi r^2=\left(\frac{\pi \rho_0 R^3}{3 \epsilon_0}\right)
$$
$\therefore \quad$ Electric field outside the ball, $E=\frac{\rho_0 R^3}{12 \in_0 r^2}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
12 balls are distributed among 3 boxes, then the probability that the first box will contain 3 balls is

Options:
A 12C3×29312
B 12C3×29310
C C312312
D C312310

The correct answer is:
12C3×29312

Solution:
Here, since each ball can be placed in any one of the 3 boxes, therefore there are 3 ways in which a ball can be placed in any one of three boxes.

Thus, there are 312 ways in which 12 balls can be placed in 3 boxes.

The number of ways in which 3 balls out of 12 can be put in the box is C312â.

The remaining 9 balls can be placed in 2 boxes in 29 ways.

Hence, the required probability is C3×2912312.

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 1)
$12 \mathrm{~g}$ of a nonvolatile solute dissolved in $108 \mathrm{~g}$ of water produces the relative lowering of vapour pressure of $0.1$. The molecular mass of the solute is :

Options:
A
$80$

B
$60$

C
$20$

D
$40$

The correct answer is:
$20$

Solution:
$\frac{\mathrm{P}^{\mathrm{o}}-\mathrm{P}_s}{\mathrm{P}^{\mathrm{o}}}=\frac{n}{N}=\frac{w}{m} \times \frac{\mathrm{M}}{W}$
$0.1=\frac{12}{m} \times \frac{18}{108}$
$m=\frac{12 \times 18}{0.1 \times 108}=20$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2013 (09 Apr Online)
\(12 \mathrm{mC}\) and \(6 \mathrm{mC}\) charges are given to the two conducting plates having same cross-sectional area and placed face to face close to each other as shown in the figure. The resulting charge distribution in \(\mathrm{mC}\) on surface \(A, B, C\) and \(D\) are respectively,

Options:
A \(9,3,-3,9\)
B \(3,9,-9,3\)
C \(6,6,-6,12\)
D \(6,6,3,3\)

The correct answer is:
\(9,3,-3,9\)

Solution:
Hint :

Topic: Physics, Capacitance, JEE Main
12 people are asked questions in succession in a random order and exactly 3 out of 12 people know the answer. The probability that the 6th person asked is the 2nd person to know the answer, is

Options:
A 1021
B

322

C 711
D 512

The correct answer is:

322

Solution:
6th Person asked is the 2nd person to know the answer

⇒In the first five people there must be exactly one person who knows the answer
Required probability =9C4×3C112C5×27
=9×8×7×6×120×324×12×11×10×9×8×27=322

Topic: Mathematics, Probability, JEE Main
1,3-Pentadiene and 1,4-pentadiene are compared with respect to their intrinsic stability and reaction with HI . The correct statement is

Options:
A 1,3-pentadiene is more stable and more reactive than 1,4-pentadiene
B 1,3-pentadiene is less stable and less reactive than 1,4-pentadiene
C 1,3-pentadiene is more stable but less reactive than 1,4-pentadiene
D 1,3-pentadiene is less stable but more reactive than 1,4-pentadiene

The correct answer is:
1,3-pentadiene is more stable and more reactive than 1,4-pentadiene

Solution:
As conjugated alkenes are more stable and more reactive due to effective resonance so 1,3-pentadiene is more stable and more reactive than 1,4-pentadiene

Topic: Chemistry, Hydrocarbons, JEE Main
1.4 g of an organic compound was digested according to Kjeldahl's method and the ammonia evolved was absorbed in 60 mL of M/10 H2SO4 solution. The excess sulphuric acid required 20 mL of M/10 NaOH solution for neutralization. The percentage of nitrogen in the compound is:

Options:
A 24
B 3
C 5
D 10

The correct answer is:
10

Solution:
Total millimoles of sulphuric acid =60×110=6 mm H2SO4
Excess H2SO4≡20×110×12=1 mm H2SO4
H2SO4 used in reaction with ammonia =6-1=5 mm
Based on the reaction stoichiometry, 2NH3+H2SO4⟶NH42SO4

mm of NH3=10 mm

Mass of N=10×10-3×14gmole=0.140g

% N2=0.1401.4×100=10%

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2015 (10 Apr Online)
15 coins are tossed. If the probability of getting at least 8 heads is equal to p, then 8p is equal to

Options:
A 2
B 4
C 8
D 16

The correct answer is:
16

Solution:
Probability of getting a head =12
Probability of getting at least 8 heads
=15C8128127+15C9129126+……+15C151215
=15C8+15C9+--+15C15215=214215=12

Topic: Mathematics, Probability, JEE Main
1.5 gm sample of bleaching powder was suspended in water. It was treated with CH3COOH followed by the addition of excess of KI. The liberated iodine required 150 mL of M10 hypo solution for complete titration. The percentage of available chlorine in the sample is

Options:

The correct answer is:
35.5

Solution:
2KI+Cl2→2KCl+I2
I2+2S2O32-→2I-+S4O62-
Mol of available Cl2≡ mole of iodine
=1×150×10-310×2
Amount of available Cl2
=150×10-320×71g
% of available Cl2
=150×10-3×71×10020×1.5=35.5

Topic: Chemistry, Redox Reactions, JEE Main
15. If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are unit coplanar vectors then the scalar triple product $[2 \mathbf{a}-\mathbf{b} 2 \mathbf{b}-\mathbf{c} 2 \mathbf{c}-\mathbf{a}]$ is equal to

Options:
A 0
B 1
C $-\sqrt{3}$
D $\sqrt{3}$

The correct answer is:
0

Solution:
Here $[\mathbf{a} \mathbf{b} \mathbf{c}]=0$
The given scalar triple product $=k[\mathbf{a b} \mathbf{c}]=0$.

Topic: Mathematics, Vector Algebra, JEE Main
16 g of a radio active substance is reduced to 0.5 g after 1 hour. The half life of the radioactive substance in minutes is

Options:

The correct answer is:
12

Solution:
Î»=1tln160.5
Î»=5Ãln2
t12=ln2Î»=15 hr=12 min

Topic: Chemistry, Chemical Kinetics, JEE Main
$0.16 \mathrm{~g}$ of an organic compound containing sulphur produces $0.233 \mathrm{~g}$ of $\mathrm{BaSO}_4$. Percentage of sulphur in the compound is

Options:
A 20
B 80
C 50
D 10

The correct answer is:
20

Solution:
Percentage of S
$\begin{aligned}
& =\frac{32 \times \text { weight of } \mathrm{BaSO}_4 \times 100}{233 \times \text { weight of compound }} \\
& =\frac{32 \times 0.233 \times 100}{233 \times 0.16} \\
& =20
\end{aligned}$

Topic: Chemistry, General Organic Chemistry, JEE Main
1.6 g of an organic compound gave 2.6 g of magnesium pyrophosphate. The percentage of phosphorus in the compound is

Options:
A 45.38%
B 54.38%
C 37.76%
D 19.02%

The correct answer is:
45.38%

Solution:
MassâofâPhosphorus =31u
Sinceâ1âmoleâof âMg2P2O7 âhas â2 gâatomsâofâP
MassâofâPhosphorusâin âMg2P2O7=62u
Massâof âMg2P2O7=222.5u
Weightâof âMg2P2O7=2.6âgâ
Weightâofâorganicâcompound =1.6âg
Percentageâofâphsphorus=Mass of phsphorus Mass of Mg2P2O7ÃWeight of Mg2P2O7 Weight of organic compoundÃ100
Percentageâofâphsphorus=62222.5Ã2.61.6Ã100=45.38%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
$16 \mathrm{~g}$ of oxygen and $3 \mathrm{~g}$ of hydrogen are mixed and kept at $760 \mathrm{~mm}$ pressure and $0^{\circ} \mathrm{C}$. The total volume occupied by the mixture will be nearly

Options:
A 22.4 litres
B 33.6 litres
C 448 litres
D $44800 \mathrm{~ml}$

The correct answer is:
$44800 \mathrm{~ml}$

Solution:
$n$ of $O_2=\frac{16}{32}=\frac{1}{2}$
n of $\mathrm{H}_2=\frac{3}{2}$
Total no. of moles $=\frac{3}{2}+\frac{1}{2}=2$
$V=\frac{n R T}{P}=\frac{2 \times .082 \times 273}{1}=44.8 \mathrm{lit}=44800 \mathrm{ml}$

Topic: Chemistry, States of Matter, JEE Main
16sin20°sin40°sin80° is equal to

Options:
A 3
B 23
C 3
D 43

The correct answer is:
23

Solution:
We know that sinθsin60-θsin60+θ=14sin3θ

Now given 16sin20°sin40°sin80°

Comparing with above formula θ=20°

we get 16sin20°sin40°sin80°=16×14×sin3×20°

=16×14×sin60° =4×32=23

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2022 (26 Jun Shift 2)
18 g glucose C6H12O6 is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution is:

Options:
A 752.4
B 759.0
C 7.6
D 76.0

The correct answer is:
752.4

Solution:
∆PPo=Mole fraction of glucose.
760-PSoln760=W1Mwt1W1M.wt1+w2M.wt2 =1818018180+178.218=0.10.1+9.9=1100
760-PSoln=760100
PSol=752.4

Topic: Chemistry, Solutions, JEE Main, JEE Main 2016 (03 Apr)
$18 \mathrm{~g}$ of glucose $\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)$ is added to $178.2 \mathrm{~g}$ of water. The vapour pressure of water for this aqueous solution at $100^{\circ} \mathrm{C}$ is

Options:
A
$759.00$ Torr

B
$7.60$ Torr

C
$76.00$ Torr

D
$752.40$ Torr

The correct answer is:
$752.40$ Torr

Solution:
$$
\begin{aligned}
& \frac{\mathrm{P}^{\circ}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{\mathrm{s}}}=\frac{\mathrm{n}}{\mathrm{N}} \\
& \frac{760-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{\mathrm{s}}}=\frac{\frac{18}{180}}{\frac{178.2}{18}}=\frac{\frac{1}{10}}{9.9}=\frac{0.1}{9.9} \\
& 760-\mathrm{P}_{\mathrm{s}}=\frac{1}{99} \mathrm{P}_{\mathrm{s}} \\
& 760 \times 99-\mathrm{P}_{\mathrm{s}} \times 99=\mathrm{P}_{\mathrm{s}} \\
& 760 \times 99=100 \mathrm{P}_{\mathrm{s}} \\
& \mathrm{P}_{\mathrm{s}}=\frac{760 \times 99}{100}=752.4
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main
$18 \mathrm{~g}$ of glucose $\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)$ is added to $178.2 \mathrm{~g}$ of water. The vapour pressure of water for this aqueous solution at $100^{\circ} \mathrm{C}$ is

Options:
A
$759.00$ Torr

B
$7.60$ Torr

C
$76.00$ Torr

D
$752.40$ Torr

The correct answer is:
$752.40$ Torr

Solution:
$$
\begin{aligned}
& \frac{\mathrm{P}^{\circ}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{\mathrm{s}}}=\frac{\mathrm{n}}{\mathrm{N}} \\
& \frac{760-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_{\mathrm{s}}}=\frac{\frac{18}{180}}{\frac{178.2}{18}}=\frac{\frac{1}{10}}{9.9}=\frac{0.1}{9.9} \\
& 760-\mathrm{P}_{\mathrm{s}}=\frac{1}{99} \mathrm{P}_{\mathrm{s}} \\
& 760 \times 99-\mathrm{P}_{\mathrm{s}} \times 99=\mathrm{P}_{\mathrm{s}} \\
& 760 \times 99=100 \mathrm{P}_{\mathrm{s}} \\
& \mathrm{P}_{\mathrm{s}}=\frac{760 \times 99}{100}=752.4
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2006
18 glucose C6H12O6 is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution at 273 K is

Options:
A 759.0
B 739.6
C 746.0
D 752.4

The correct answer is:
752.4

Solution:
Relative lowering of vapour pressure
Equation P0-PsP0=Xsolute=nn+N
Modified forms of equation is P0-PsPs=nN
n = moles of solute, N = moles of solvent
P0=760 torr
Ps=?
760-PsPs=18180178.218
Ps=752.4 torr

Topic: Chemistry, Solutions, JEE Main
18 gram glucose (Molar mass = 180) is dissolved in 100 mol of water at 300 K. If R = 3 0.0821 L-atm mol-1K-1 what is the osmotic pressure of solution?

Options:
A 2.463 atm
B 24.63 atm
C 8.21 atm
D 0.821 atm

The correct answer is:
24.63 atm

Solution:
The various quantities known to us are as follows:
R=0.0821 L-atm mol-1K-1
w2=18 gram
Molar mass M2=180
T=300 K
V=100 mL
To calculate the osmotic pressure of solution, we use the following formula,
π=w2RTM2V=18×0.0821×300×1000180×100
=24.63 atm

Topic: Chemistry, Solutions, JEE Main
$19 \mathrm{~g}$ of a mixture containing $\mathrm{NaHCO}_3$ and $\mathrm{Na}_2 \mathrm{CO}_3$ on complete heating liberated $1.12 \mathrm{~L}$ of $\mathrm{CO}_2$ at STP. The weight of the remaining solid was $15.9 \mathrm{~g}$. What is the weight (in g) of $\mathrm{Na}_2 \mathrm{CO}_3$ in the mixture before heating?

Options:
A 8.4
B 15.9
C 4.0
D 10.6

The correct answer is:
10.6

Solution:
Molecular weight of
$\mathrm{NaHCO}_3=23+1+12+48=84$
Molecular weight of
$\mathrm{Na}_2 \mathrm{CO}_3=46+12+48=106$
Hence, total weight $=84+106=190$
$\because$ In $190 \mathrm{~g}$ of a mixture, weight of $\mathrm{Na}_2 \mathrm{CO}_3$ is
$=106$
$\therefore$ In $19 \mathrm{~g}$ of a mixture weight of
$\begin{aligned}
\mathrm{Na}_2 \mathrm{CO}_3 & =\frac{106 \times 19}{190} \\
& =10.6 \mathrm{~g}
\end{aligned}$

Topic: Chemistry, Some Basic Concepts of Chemistry, AP EAMCET, AP EAMCET 2011
$19 \mathrm{~g}$ of a mixture containing $\mathrm{NaHCO}_3$ and $\mathrm{Na}_2 \mathrm{CO}_3$ on complete heating liberated $1.12 \mathrm{~L}$ of $\mathrm{CO}_2$ at STP. The weight of the remaining solid was $15.9 \mathrm{~g}$. What is the weight (in g) of $\mathrm{Na}_2 \mathrm{CO}_3$ in the mixture before heating?

Options:
A 8.4
B 15.9
C 4.0
D 10.6

The correct answer is:
10.6

Solution:
Molecular weight of
$\mathrm{NaHCO}_3=23+1+12+48=84$
Molecular weight of
$\mathrm{Na}_2 \mathrm{CO}_3=46+12+48=106$
Hence, total weight $=84+106=190$
$\because$ In $190 \mathrm{~g}$ of a mixture, weight of $\mathrm{Na}_2 \mathrm{CO}_3$ is
$=106$
$\therefore$ In $19 \mathrm{~g}$ of a mixture weight of
$\begin{aligned}
\mathrm{Na}_2 \mathrm{CO}_3 & =\frac{106 \times 19}{190} \\
& =10.6 \mathrm{~g}
\end{aligned}$

Topic: Chemistry, Some Basic Concepts of Chemistry, TS EAMCET, TS EAMCET 2011
20 ml of 0.1 M NaOH is added to 30 ml0.1 M NaOH 30 mlâ0.2 M CH 3 COOH pK a =4.74 , calculate the pH of the resulting solution? [given: logâ¡2=0.3 ]

Options:
A 3.44
B 4.01
C 4.44
D 4.71

The correct answer is:
4.44

Solution:
NaOH+CH3COOHâCH3COONa+H2Om mol2 6 LR - 4 2
The resulting solution is an acidic buffer. so pH= pK a +log Salt Acid
=4.71+logâ¡24
=4.44

Topic: Chemistry, Ionic Equilibrium, JEE Main
$20 \mathrm{ml}$ of $0.1 \mathrm{M}$ acetic acid is mixed with $50 \mathrm{~mL}$ of potassium acetate. $\mathrm{K}_{\mathrm{a}}$ of acetic acid $=1.8 \times 10^{-5}$ at $27^{\circ} \mathrm{C} .$ Calculate concentration of potassium acetate if $\mathrm{pH}$ of the mixture is $4.8$.

Options:
A $0.1 \mathrm{M}$
B $0.04 \mathrm{M}$
C $0.4 \mathrm{M}$
D $0.02 \mathrm{M}$

The correct answer is:
$0.1 \mathrm{M}$

Solution:
Let the concentration of potassium acetate is $x$. According to Henderson's equation,
$$
\begin{array}{l}
\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\mathrm{salt}]}{[\mathrm{acid}]} \\
4.8=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{x \times 50}{20 \times 0.1 \mathrm{M}} \\
4.8=4.74+\log 25 x \\
\text { or } \log 25 x=0.06
\end{array}
$$
$$
\begin{array}{l}
25 x=1.148 \\
\therefore x=0.045 \mathrm{M}
\end{array}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main
20 ml of 0.1 M H2SO4  solution is added to 30 mL of 0.2 M NH4OH solution. The pH of the resultant mixture is: pkb of NH4OH=4.7

Options:
A 5.0
B 5.2
C 9.4
D 9.0

The correct answer is:
9.0

Solution:
Millimoles of H2SO4=20×0.1=2
Millimoles of NH4OH=30×0.2=6
Hence, the solution will act as a basic buffer.
pOH=pKb+logsaltBase
NH4+=2×NH42SO4; NH4+=4
pOH=pKb+logNH4+NH3
=4.7+log42
=4.7+0.3010=5.010
pH=14-pOH=9

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2019 (09 Jan Shift 1)
$20 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCl}$ is added to the $30 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{NaOH}$. To this solution extra $50 \mathrm{~mL}$ of water was added. What is the molarity of the final solution formed?

Options:
A $0.1 \mathrm{M}$
B $0.01 \mathrm{M}$
C $0.5 \mathrm{M}$
D $0.05 \mathrm{M}$

The correct answer is:
$0.01 \mathrm{M}$

Solution:
$\mathrm{n}_{\mathrm{HCl}}=0.1 \mathrm{M} \times 0.02 \mathrm{~L}=0.002 \mathrm{~mol}$
$\mathrm{n}_{\mathrm{NaOH}}=0.1 \mathrm{M} \times 0.03 \mathrm{~L}=0.003 \mathrm{~mol}$
Number of moles of $\mathrm{NaOH}$ remaining after neutralization. $=0.003-0.002=0.001 \mathrm{~mol}$.
Total final volume $=50 \mathrm{~mL}+50 \mathrm{~mL}=100 \mathrm{~mL}=0.1 \mathrm{~L}$ $\Rightarrow$ Molarity $=\frac{0.001}{0.1}=0.01 \mathrm{M}$

Topic: Chemistry, Some Basic Concepts of Chemistry, AP EAMCET, AP EAMCET 2023 (15 May Shift 1)