New chat PYQ Library Doubts Rank report

Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →

Solutions question

2023 · 31 Jan · Shift 1 · Q20

Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.

PYQ Library
JEE Main
Chemistry
Solutions
2023 · 31 Jan · Shift 1 · Q20

Solutions question

2023 · 31 Jan · Shift 1 · Q20

JEE MainChemistrySolutionsNumerical+4 / −1

At

27^{\circ} \mathrm{C}

, a solution containing

2.5 \mathrm{~g}

of solute in

250.0 \mathrm{~mL}

of solution exerts an osmotic pressure of

400 \mathrm{~Pa}

. The molar mass of the solute is

\underline{\hspace{2cm}}\mathrm{g} \mathrm{~mol}^{-1}

(Nearest integer) (Given :

\mathrm{R}=0.083 \mathrm{~L} \mathrm{~bar} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

)

Numerical answer

View written solutionFree

Correct answer: 62250

Use the osmotic pressure relation

For a dilute solution,

\pi = CRT = \frac{n}{V}RT

where:

$\pi$ = osmotic pressure
$C$ = molarity
$n$ = moles of solute
$V$ = volume of solution
$R$ = gas constant
$T$ = temperature

Also,

n = \frac{w}{M}

where $w$ is mass of solute and $M$ is molar mass.

So,

\pi = \frac{w}{MV}RT

Hence,

M = \frac{wRT}{\pi V}

Convert all quantities into consistent units

Given:

$w = 2.5\,\text{g}$
$V = 250.0\,\text{mL} = 0.250\,\text{L}$
$T = 27^\circ C = 300\,\text{K}$
$R = 0.083\,\text{L bar K}^{-1}\text{mol}^{-1}$
$\pi = 400\,\text{Pa}$

Since $R$ is in $\text{L bar K}^{-1}\text{mol}^{-1}$ , convert pressure from Pa to bar:

1\,\text{bar} = 10^5\,\text{Pa}

So,

400\,\text{Pa} = \frac{400}{10^5}\,\text{bar} = 4\times 10^{-3}\,\text{bar}

Substitute into the formula

M = \frac{(2.5)(0.083)(300)}{(4\times 10^{-3})(0.250)}

First, numerator:

2.5 \times 0.083 \times 300 = 62.25

Denominator:

4\times 10^{-3} \times 0.250 = 0.001

Therefore,

M = \frac{62.25}{0.001} = 62250\,\text{g mol}^{-1}

Nearest integer

\boxed{62250}

Comparison with stored correct answer

Stored correct answer = $62250$

This matches exactly.

More from Solutions

Evaluate the following statements for their correctness. A. The elevation in boiling point temperature of water will be same for $0.1 M NaCl$ and $0.1 M$ urea. B. Azeotropic mixtures boil without change in…2023 · MCQ
The osmotic pressure of blood is 7.47 bar at 300 K. To inject glucose to a patient intravenously, it has to be isotonic with blood. The concentration of glucose solution in gL $-$ 1 is . (Molar mass of glucose =…2022 · Numerical
A company dissolves 'x' amount of $C O_{2}$ at 298 K in 1 litre of water to prepare soda water. X = $\times$ 10 $-$ 3 g. (nearest integer) (Given : partial pressure of $C O_{2}$ at 298 K = 0.835 bar. Henry's law constant…2022 · Numerical
The depression in freezing point observed for a formic acid solution of concentration $0.5 mL L^{- 1}$ is $0.040 5^{\circ} C$ . Density of formic acid is $1.05 g mL^{- 1}$ . The Van't Hoff factor…2022 · MCQ
Two solutions A and B are prepared by dissolving 1 g of non-volatile solutes X and Y, respectively in 1 kg of water. The ratio of depression in freezing points for A and B is found to be 1 : 4. The ratio of molar masses of X and Y is2022 · MCQ
Solute A associates in water. When 0.7 g of solute A is dissolved in 42.0 g of water, it depresses the freezing point by 0.2 $^{\circ}$ C. The percentage association of solute A in water, is : [Given : Molar mass of A = 93 g mol $-$ 1. Molal…2022 · MCQ
The elevation in boiling point for 1 molal solution of non-volatile solute A is $A$ . The depression in freezing point for 2 molal solution of $A$ in the same solvent is 6 $K$ . The ratio of $K_{b}$ and $K_{f}$ i.e., $K_{b} / K_{f}$ …2022 · Numerical
A 0.5 percent solution of potassium chloride was found to freeze at $-$ 0.24 $^{\circ}$ C. The percentage dissociation of potassium chloride is . (Nearest integer) (Molal depression constant for water is 1.80 K kg…2022 · Numerical