datavorous/entrance-exam-dataset huggingface dataset:
Paragraph:
Carbon-14 is used to determine the age of organic material. The procedure is based on the formation of ${ }^{14} \mathrm{C}$ by neutron capture in the upper atmosphere.
$$
{ }_7^{14} \mathrm{~N}+{ }_0^1 n \longrightarrow{ }_6^{14} \mathrm{C}+{ }_1 n^1
$$
${ }^{14} \mathrm{C}$ is absorbed by living organisms during photosynthesis. The ${ }^{14} \mathrm{C}$ content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of ${ }^{14} \mathrm{C}$ in the dead being, falls to the decay which ${ }^{14} \mathrm{C}$ undergoes.
$$
{ }_6^{14} \mathrm{C} \longrightarrow{ }_7^{14} \mathrm{~N}+\beta^{-}
$$
The half life period of ${ }^{14} \mathrm{C}$ is 5770 years. The decay constant $(\lambda)$ can be calculated by using the following formula $\lambda=\frac{0.693}{t_{1 / 2}}$.
The comparison of the $\beta^{-}$activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of ${ }^{14} \mathrm{C}$ to ${ }^{12} \mathrm{C}$ in living matter is $1: 10^{12}$.Question:
Which of the following option is correct?

Options:
A
In living organisms, circulation of ${ }^{14} \mathrm{C}$ from atmosphere is high so the carbon content is constant in organism.

B
Carbon dating can be used to find out the age of earth crust and rocks.

C
Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbon content remains constant in living organisms.

D
Carbon dating can not be used to determine concentration of ${ }^{14} \mathrm{C}$ in dead beings.

The correct answer is:
Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbon content remains constant in living organisms.

Solution:
Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbon content as the ratio of $\mathrm{C}^{14}$ to $\mathrm{C}^{12}$ remains constant in living organism.

Topic: Chemistry, Chemical Kinetics, JEE Main
Paragraph:
Carbon-14 is used to determine the age of organic material. The procedure is based on the formation of ${ }^{14} \mathrm{C}$ by neutron capture in the upper atmosphere.
$$
{ }_7^{14} \mathrm{~N}+{ }_0^1 n \longrightarrow{ }_6^{14} \mathrm{C}+{ }_1 n^1
$$
${ }^{14} \mathrm{C}$ is absorbed by living organisms during photosynthesis. The ${ }^{14} \mathrm{C}$ content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of ${ }^{14} \mathrm{C}$ in the dead being, falls to the decay which ${ }^{14} \mathrm{C}$ undergoes.
$$
{ }_6^{14} \mathrm{C} \longrightarrow{ }_7^{14} \mathrm{~N}+\beta^{-}
$$
The half life period of ${ }^{14} \mathrm{C}$ is 5770 years. The decay constant $(\lambda)$ can be calculated by using the following formula $\lambda=\frac{0.693}{t_{1 / 2}}$.
The comparison of the $\beta^{-}$activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of ${ }^{14} \mathrm{C}$ to ${ }^{12} \mathrm{C}$ in living matter is $1: 10^{12}$.Question:
A nuclear explosion has taken place leading to increase in concentration of $C^{14}$ in nearby areas. $\mathrm{C}^{14}$ concentration is $C_1$ in nearby areas and $C_2$ in areas far away. If the age of the fossil is determined to be $T_1$ and $T_2$ at the places respectively then

Options:
A
The age of the fossil will increase at the place where explosion has taken and
$$
T_1-T_2=\frac{1}{\lambda} \ln \frac{C_1}{C_2}
$$

B
The age of the fossil will decrease at the place where explosion has taken place and $T_1-T_2=\frac{1}{\lambda} \ln \frac{C_1}{C_2}$

C
The age of fossil will be determined to be same

D
$\frac{T_1}{T_2}=\frac{C_1}{C_2}$

The correct answer is:
The age of the fossil will increase at the place where explosion has taken and
$$
T_1-T_2=\frac{1}{\lambda} \ln \frac{C_1}{C_2}
$$

Solution:
All radioactive decays are the examples of first order kinetics.
So, decay constant $\lambda=\frac{1}{T_1-T_2} \ln \frac{C_1}{C_2}$
$C_1$ is the concentration at $T_1$ time
$C_2$ is the concentration at $T_2$ time
So, $\quad T_1-T_2=\frac{1}{\lambda} \ln \frac{C_1}{C_2}$

Topic: Chemistry, Chemical Kinetics, JEE Main
Paragraph:
Consider a block of conducting material of resistivity ' $\rho$ ' shown in the figure. Current 'l' enters at 'A' and leaves from ' $\mathrm{D}$ '. We apply superposition principle to find voltage ' $\Delta \mathrm{V}$ ' developed between ' $\mathrm{B}$ ' and ' $\mathrm{C}$ '. The calculation is done in the following steps:
(i) Take current 'l' entering from 'A' and assume it to spread over a hemispherical surface in the block.
(ii) Calculate field $E(r)$ at distance ' $r$ ' from $A$ by using Ohm's law $E=\rho j$, where $j$ is the current per unit area at ' $r$ '.
(iii) From the ' $r$ ' dependence of $E(r)$, obtain the potential $V(r)$ at $r$.
(iv) Repeat (i), (ii) and (iii) for current 'l' leaving ' $D$ ' and superpose results for ' $A$ ' and ' $D$ '.
Question:
$\Delta \mathrm{V}$ measured between $\mathrm{B}$ and $\mathrm{C}$ is

Options:
A
$\frac{\rho l}{\pi a}-\frac{\rho l}{\pi(a+b)}$

B
$\frac{\rho l}{a}-\frac{\rho l}{(a+b)}$

C
$\frac{\rho l}{2 \pi a}-\frac{\rho l}{2 \pi(a+b)}$

D
$\frac{\rho l}{2 \pi(a-b)}$

The correct answer is:
$\frac{\rho l}{2 \pi a}-\frac{\rho l}{2 \pi(a+b)}$

Solution:
Choosing $A$ as origin,
$$
\begin{gathered}
E=\rho j=\rho \frac{1}{2 \pi r^2} \\
V_C-V_B=-\frac{\rho l}{2 \pi} \int_a^{(a+b)} \frac{1}{r^2} d r=\frac{\rho l}{2 \pi}\left[\frac{1}{(a+b)}-\frac{1}{a}\right] \\
V_B-V_C=\frac{\rho l}{2 \pi}\left[\frac{1}{a}-\frac{1}{(a+b)}\right]
\end{gathered}
$$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2008
Paragraph:
Consider a block of conducting material of resistivity ' $\rho$ ' shown in the figure. Current 'l' enters at 'A' and leaves from ' $\mathrm{D}$ '. We apply superposition principle to find voltage ' $\Delta \mathrm{V}$ ' developed between ' $\mathrm{B}$ ' and ' $\mathrm{C}$ '. The calculation is done in the following steps:
(i) Take current 'l' entering from 'A' and assume it to spread over a hemispherical surface in the block.
(ii) Calculate field $E(r)$ at distance ' $r$ ' from $A$ by using Ohm's law $E=\rho j$, where $j$ is the current per unit area at ' $r$ '.
(iii) From the ' $r$ ' dependence of $E(r)$, obtain the potential $V(r)$ at $r$.
(iv) Repeat (i), (ii) and (iii) for current 'l' leaving ' $D$ ' and superpose results for ' $A$ ' and ' $D$ '.
Question:
For current entering at $A$, the electric field at a distance ' $r$ ' from $A$ is

Options:
A
$\frac{\rho l}{8 \pi r^2}$

B
$\frac{\rho l}{r^2}$

C
$\frac{\rho l}{2 \pi r^2}$

D
$\frac{\rho l}{4 \pi r^2}$

The correct answer is:
$\frac{\rho l}{2 \pi r^2}$

Solution:
No Solution Available

Topic: Physics, Current Electricity, JEE Main, JEE Main 2008
Paragraph:
Consider the functions defined implicitly by the equation $y^3-3 y+x=0$ on various intervals in the real line. If $x \in(-\infty,-2) \cup(2, \infty)$, the equation implicitly defines a unique real valued differentiable function $y=f(x)$. If $x \in(-2,2)$, the equation implicitly defines a unique real valued differentiable function $y=g(x)$, satisfying $g(0)=0$.Question:
If $f(-10 \sqrt{2})=2 \sqrt{2}$, then $f^{\prime \prime}(-10 \sqrt{2})$ is equal to

Options:
A
$\frac{4 \sqrt{2}}{7^3 \cdot 3^2}$

B
$-\frac{4 \sqrt{2}}{7^3 \cdot 3^2}$

C
$\frac{4 \sqrt{2}}{7^3 \cdot 3}$

D
$-\frac{4 \sqrt{2}}{7^3 \cdot 3}$

The correct answer is:
$-\frac{4 \sqrt{2}}{7^3 \cdot 3^2}$

Solution:
Given, $y^3-3 y+x=0 \Rightarrow 3 y^2 \frac{d y}{d x}-3 \frac{d y}{d x}+1=0$
$$
\Rightarrow \quad 3 y^2\left(\frac{d^2 y}{d x^2}\right)+6 y\left(\frac{d y}{d x}\right)^2-3 \frac{d^2 y}{d x^2}=0
$$
On substituting $x=-10 \sqrt{2}, y=2 \sqrt{2}$ in Eq. (i), we get
$$
3(2 \sqrt{2})^2 \cdot \frac{d y}{d x}-3 \cdot \frac{d y}{d x}+1=0 \Rightarrow \frac{d y}{d x}=\frac{-1}{21}
$$
Again on substituting $x=-10 \sqrt{2}, y=2 \sqrt{2}$ in Eq. (ii), we get
$$
\begin{aligned}
& 3(2 \sqrt{2})^2 \frac{d^2 y}{d x^2}+6(2 \sqrt{2}) \cdot\left(\frac{-1}{21}\right)^2-3 \cdot \frac{d^2 y}{d x^2}=0 \\
& \Rightarrow \quad 21 \cdot \frac{d^2 y}{d x^2}=-\frac{12 \sqrt{2}}{(21)^2} \Rightarrow \frac{d^2 y}{d x^2}=-\frac{12 \sqrt{2}}{(21)^3}=-\frac{4 \sqrt{2}}{7^3 \cdot 3^2}
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
Paragraph:
Consider the functions defined implicitly by the equation $y^3-3 y+x=0$ on various intervals in the real line. If $x \in(-\infty,-2) \cup(2, \infty)$, the equation implicitly defines a unique real valued differentiable function $y=f(x)$. If $x \in(-2,2)$, the equation implicitly defines a unique real valued differentiable function $y=g(x)$, satisfying $g(0)=0$.Question:
The area of the region bounded by the curve $y=f(x)$, the $X$-axis and the lines $x=a$ and $x=b$, where $-\infty < a < b < -2$, is

Options:
A
$\int_a^b \frac{x}{3\left[f(x)^2-1\right]} d x+b f(b)-a f(a)$

B
$-\int_a^b \frac{x}{3\left[(f(x))^2-1\right]} d x+b f(b)-a f(a)$

C
$\int_a^b \frac{x}{3\left[(f(x))^2-1\right]} d x-b f(b)+a f(a)$

D
$-\int_a^b \frac{x}{3\left[f(x)^2-1\right]} d x-b f(b)-a f(a)$

The correct answer is:
$\int_a^b \frac{x}{3\left[f(x)^2-1\right]} d x+b f(b)-a f(a)$

Solution:
Required area $=\int_a^b y d x=\int_a^b f(x) d x$
$$
=[f(x) \cdot x]_a^b-\int_a^b f^{\prime}(x) \cdot x d x=b f(b)-a f(a)-\int_a^b f^{\prime}(x) \cdot x d x
$$
$$
=b f(b)-a f(a)+\int_a^b \frac{x}{3\left[(f(x))^2-1\right]} d x
$$
As, $\quad f^{\prime}(x)=\frac{d y}{d x}=-\frac{1}{3\left(y^2-1\right)}=-\frac{1}{3\left[(f(x))^2-1\right]}$

Topic: Mathematics, Area Under Curves, JEE Main
Paragraph:
If a continuous $f$ defined on the real line $R$, assume positive and negative values in $R$, then the equation $f(x)=0$ has a root in $R$. For example, if it is known that a continuous function $f$ on $R$ is positive at some point and its minimum values is negative, then the equation $f(x)=0$ has a root in $R$.
Consider $f(x)=k e^x-x$ for all real $x$, where $k$ is real constant.Question:
The positive value of $k$ for which $k e^x-x=0$ has only one root is

Options:
A
$\frac{1}{e}$

B
1

C
$e$

D
$\log _e 2$

The correct answer is:
$\frac{1}{e}$

Solution:
Let $f(x)=k e^x-x$
$\begin{aligned} & & f^{\prime}(x) & =k e^x-1=0 \\ & & x & =-\ln k \\ & \therefore & f^{\prime \prime}(x) & =k e^x \\ & \text { Hence, } & {\left[f^{\prime \prime}(x)\right]_{x=-\ln k} } & =1>0 \\ & & f(-\ln k) & =1+\ln k\end{aligned}$
For one root of given equation
$$
\begin{aligned}
1+\ln k & =0 \\
Hence,
k & =\frac{1}{e} .
\end{aligned}
$$

Topic: Mathematics, Functions, JEE Main
Paragraph:
In a mixture of $\mathrm{H}-\mathrm{H}^{+}$gas $\left(\mathrm{He}^{+}\right.$is singly ionised $\mathrm{He}$ atom), $\mathrm{H}$ atoms and $\mathrm{He}^{+}$ions are excited to their respective first excited states. Subsequently, $\mathrm{H}$ atoms transfer their total excitation energy to $\mathrm{He}^{+}$ions (by collisions). Assume that the Bohr model of atom is exactly valid.Question:
The wavelength of light emitted in the visible region by $\mathrm{He}^{+}$ions after collisions with $\mathrm{H}$ atoms is

Options:
A
$6.5 \times 10^{-7} \mathrm{~m}$

B
$5.6 \times 10^{-7} \mathrm{~m}$

C
$4.8 \times 10^{-7} \mathrm{~m}$

D
$4.0 \times 10^{-7} \mathrm{~m}$

The correct answer is:
$4.8 \times 10^{-7} \mathrm{~m}$

Solution:
Visible light lies in the range, $\lambda_1=4000 Å$ to $\lambda_2=7000 Å$.
Energy of photons corresponding to these wavelengths (in $\mathrm{eV}$ ) would be;
$$
\begin{aligned}
& E_1=\frac{12375}{4000}=3.09 \mathrm{eV} \\
& E_2=\frac{12375}{7000}=1.77 \mathrm{eV}
\end{aligned}
$$
From energy level diagram of $\mathrm{He}^{+}$atom we can see that in transition from $n=4$ to $n=3$, energy of photon released will lie between $E_1$ and $E_2$.
$$
\Delta E_{43}=-3.4-(-6.04)=2.64 \mathrm{eV}
$$
Wavelength of photon corresponding to this energy.
$$
\lambda=\frac{12375}{2.64} Å=4687.5 Å=4.68 \times 10^{-7} \mathrm{~m}
$$
Therefore, (c) is the most correct option.

Topic: Physics, Atomic Physics, JEE Main
Paragraph:
Let $p$ be an odd prime number and $T_p$ be the following set of $2 \times 2$ matrices
$$
T_p=\left\{A=\left[\begin{array}{ll}
a & b \\
c & a
\end{array}\right] ; a, b, c \in\{0,1,2, \ldots, p-1\}\right\}
$$Question:
The number of $A$ in $T_p$ such that $A$ is either symmetric or skew-symmetric or both, and det $(A)$ is divisible by $p$ is

Options:
A
$(p-1)^2$

B
$2(p-1)$

C
$(p-1)^2+1$

D
$2 p-1$

The correct answer is:
$2 p-1$

Solution:
Given, $A=\left[\begin{array}{ll}a & b \\ c & a\end{array}\right]$,
$a, b, c \in\{0,1,2, \ldots, p-1\}$
If $A$ is skew-symmetric matrix, then $a=0, b=-c$
$\therefore \quad|A|=-b^2$.
Thus, $P$ divides $|A|$ only when $b=0$...(i)
Again, if $A$ is symmetric matrix, then $b=c$ and $|A|=a^2-b^2$.
Thus, $p$ divides $|A|$ if either $p$ divides $(a-b)$ or $p$ divides $(a+b)$. $p$ divides $(a-b)$, only when $a=b$ ie, $a=b \in\{0,1,2, \ldots,(p-1)\}$
ie, pchoices
$p$ divides $(a+b)$.
$\Rightarrow p$ choices, including $a=b=0$ included in (i)
$\therefore$ Total number of choices are $(p+p-1)=2 p-1$.
Hence, (c) is the correct option.

Topic: Mathematics, Matrices, JEE Main
Paragraph:
Let $V_r$ denotes the sum of the first $r$ terms of an arithmetic progression $(A P)$ whose first term is $r$ and the common difference is $(2 r-1)$. Let $T_r=V_{r+1}-V_r-2$ and $Q_r=T_{r+1}-T_r$ for $r=1,2, \ldots$Question:
The sum $V_1+V_2+\ldots+V_n$ is

Options:
A
$\frac{1}{12} n(n+1)\left(3 n^2-n+1\right)$

B
$\frac{1}{12} n(n+1)\left(3 n^2+n+2\right)$

C
$\frac{1}{2} n\left(2 n^2-n+1\right)$

D
$\frac{1}{3}\left(2 n^3-2 n+3\right)$

The correct answer is:
$\frac{1}{12} n(n+1)\left(3 n^2+n+2\right)$

Solution:
$$
\begin{aligned}
& V_r=\frac{r}{2}[2 r+(r-1)(2 r-1)]=\frac{1}{2}\left(2 r^3-r^2+r\right) \\
& \therefore \quad \Sigma V_r=\frac{1}{2}\left[2 \Sigma r^3-\Sigma r^2+\Sigma r\right] \\
& =\frac{1}{2}\left[2\left(\frac{n(n+1)}{2}\right)^2-\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\right] \\
& =\frac{1}{12} n(n+1)\left(3 n^2+n+2\right) \text {. } \\
&
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main
Paragraph:
$\mathbf{P}_{17 \text { - } 19}$ : Paragraph for Questions Nos. 17 to 19 Two discs $A$ and $B$ are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I, respectively about the common axis. Disc $A$ is imparted an initial angular velocity $2 \omega$ using the entire potential energy of a spring compressed by a distance $x_1$. Disc $B$ is imparted an angular velocity $\omega$ by a spring having the same spring constant and compressed by a distance $x_2$. Both the discs rotate in the clockwise direction.Question:
The ratio $\frac{x_1}{x_2}$ is

Options:
A
2

B
$\frac{1}{2}$

C
$\sqrt{2}$

D
$\frac{1}{\sqrt{2}}$

The correct answer is:
$\sqrt{2}$

Solution:
$$
\begin{aligned}
\frac{1}{2} I(2 \omega)^2 & =\frac{1}{2} k x_1^2 \\
\frac{1}{2}(2 I)(\omega)^2 & =\frac{1}{2} k x_2^2
\end{aligned}
$$
From Eqs. (i) and (ii), we have
$$
\frac{x_1}{x_2}=\sqrt{2}
$$
$\therefore$ Option (c) is correct.

Topic: Physics, Rotational Motion, JEE Main
Paragraph:
$\mathbf{P}_{17 \text { - } 19}$ : Paragraph for Questions Nos. 17 to 19 Two discs $A$ and $B$ are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I, respectively about the common axis. Disc $A$ is imparted an initial angular velocity $2 \omega$ using the entire potential energy of a spring compressed by a distance $x_1$. Disc $B$ is imparted an angular velocity $\omega$ by a spring having the same spring constant and compressed by a distance $x_2$. Both the discs rotate in the clockwise direction.Question:
When disc $B$ is brought in contact with disc $A$, they acquire a common angular velocity in time $t$. The average frictional torque on one disc by the other during this period is

Options:
A
$\frac{2 I \omega}{3 t}$

B
$\frac{9 I \omega}{2 t}$

C
$\frac{91 \omega}{4 t}$

D
$\frac{3 I \omega}{2 t}$

The correct answer is:
$\frac{2 I \omega}{3 t}$

Solution:
Let $\omega^{\prime}$ be the common velocity. Then from conservation of angular momentum, we have
$$
\begin{aligned}
(i+2 I) \omega^{\prime} & =I(2 \omega)+2 I(\omega) \\
\omega^{\prime} & =\frac{4}{3} \omega
\end{aligned}
$$
From the equation,
Angular impulse = change in angular momentum for any of the disc, we have
$$
\begin{aligned}
\tau t & =I(2 \omega)-I\left(\frac{4}{3} \omega\right)=\frac{2 I \omega}{3} \\
\tau & =\frac{2 I \omega}{3 t}
\end{aligned}
$$
$\therefore$ Option (a) is correct.

Topic: Physics, Rotational Motion, JEE Main
Paragraph:
$\mathbf{P}_{17 \text { - } 19}$ : Paragraph for Questions Nos. 17 to 19 Two discs $A$ and $B$ are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I, respectively about the common axis. Disc $A$ is imparted an initial angular velocity $2 \omega$ using the entire potential energy of a spring compressed by a distance $x_1$. Disc $B$ is imparted an angular velocity $\omega$ by a spring having the same spring constant and compressed by a distance $x_2$. Both the discs rotate in the clockwise direction.Question:
The loss of kinetic energy during the above process is

Options:
A
$\frac{I \omega^2}{2}$

B
$\frac{i \omega^2}{3}$

C
$\frac{I \omega^2}{4}$

D
$\frac{I \omega^2}{6}$

The correct answer is:
$\frac{i \omega^2}{3}$

Solution:
Loss of kinetic energy $=K_i-K_f$
$$
\begin{aligned}
& =\left\{\frac{1}{2} I(2 \omega)^2+\frac{1}{2}(2 I)(\omega)^2\right\}-\frac{1}{2}(3 I)\left(\frac{4}{3} \omega\right)^2 \\
& =\frac{1}{3} I \omega^2
\end{aligned}
$$
$\therefore$ Option (b) is correct.

Topic: Physics, Rotational Motion, JEE Main
Paragraph:
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator automobiles.
A solution $M$ is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixtrue is $0.9$
Given Freezing point depression constant of water $\left(k^{\text {water }}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Freezing point depression constant of ethanol $\left(k_f^{\text {ethanol }}\right)=2.0 \mathrm{Kkg} \mathrm{mol}^{-1}$
Boiling point elevation constant of water $\left(k_b^{\text {water }}\right)=0.52 \mathrm{Kkg} \mathrm{mol}^{-1}$
Boiling point elevation constant of ethanol $\left(k_b^{\text {ethanol }}\right)=1.2 \mathrm{Kkg} \mathrm{mol}^{-1}$
Standard freezing point of water $=273 \mathrm{~K}$
Standard freezing point of ethanol $=155.7 \mathrm{~K}$
Standard boiling point of water $=373 \mathrm{~K}$
Standard boiling point of ethanol $=351.5 \mathrm{~K}$
Vapour pressure of pure water $=328 \mathrm{~mm}$ of $\mathrm{Hg}$
Vapour pressure of pure ethanol $=40 \mathrm{~mm}$ of $\mathrm{Hg}$
Molecular weight of water $=18 \mathrm{~g} \mathrm{~mol}^{-1}$
Molecular weight of ethanol $=46 \mathrm{~g} \mathrm{~mol}^{-1}$
In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.Question:
The vapour pressure of the solution $M$ is

Options:
A
$39.3 \mathrm{~mm} \mathrm{Hg}$

B
$36.0 \mathrm{~mm} \mathrm{Hg}$

C
$29.5 \mathrm{~mm} \mathrm{Hg}$

D
$28.8 \mathrm{~mm} \mathrm{Hg}$

The correct answer is:
$36.0 \mathrm{~mm} \mathrm{Hg}$

Solution:
Total vapour pressure, $p=p_A^{\circ} \chi_A$
$$
p=40 \times 0.9=36 \mathrm{~mm} \text { of } \mathrm{Hg}
$$
In the paragraph, it has been directed to take solute as non-volatile, thus $\mathrm{H}_2 \mathrm{O}$ do not contribute in the total vapour pressure.

Topic: Chemistry, Solutions, JEE Main
Paragraph:
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator automobiles.
A solution $M$ is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixtrue is $0.9$
Given Freezing point depression constant of water $\left(k^{\text {water }}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Freezing point depression constant of ethanol $\left(k_f^{\text {ethanol }}\right)=2.0 \mathrm{Kkg} \mathrm{mol}^{-1}$
Boiling point elevation constant of water $\left(k_b^{\text {water }}\right)=0.52 \mathrm{Kkg} \mathrm{mol}^{-1}$
Boiling point elevation constant of ethanol $\left(k_b^{\text {ethanol }}\right)=1.2 \mathrm{Kkg} \mathrm{mol}^{-1}$
Standard freezing point of water $=273 \mathrm{~K}$
Standard freezing point of ethanol $=155.7 \mathrm{~K}$
Standard boiling point of water $=373 \mathrm{~K}$
Standard boiling point of ethanol $=351.5 \mathrm{~K}$
Vapour pressure of pure water $=328 \mathrm{~mm}$ of $\mathrm{Hg}$
Vapour pressure of pure ethanol $=40 \mathrm{~mm}$ of $\mathrm{Hg}$
Molecular weight of water $=18 \mathrm{~g} \mathrm{~mol}^{-1}$
Molecular weight of ethanol $=46 \mathrm{~g} \mathrm{~mol}^{-1}$
In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.Question:
The freezing point of the solution $M$ is

Options:
A
$268.7 \mathrm{~K}$

B
$268.5 \mathrm{~K}$

C
$234.2 \mathrm{~K}$

D
$150.9 \mathrm{~K}$

The correct answer is:
$150.9 \mathrm{~K}$

Solution:
Solution $M$ is mixture of ethanol and water.
Mole fraction of ethanol is $0.9 \Rightarrow$ Solvent is $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
Mole fraction of water is $0.1 \Rightarrow \mathrm{H}_2 \mathrm{O}$ is solute
Molality of $\mathrm{H}_2 \mathrm{O}=\frac{n_2}{n_1 M_1}=\frac{0.1}{0.9 \times 46} \times 1000=2.415$
$$
\Delta T_f=k_f \cdot m=2 \times 2.415=4.83
$$
or Freezing point of solution $=155.7-4.83=150.87 \mathrm{~K}$

Topic: Chemistry, Solutions, JEE Main