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Solutions question

2018 · 16 Apr · Shift 1 · Q21

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2018 · 16 Apr · Shift 1 · Q21

Solutions question

2018 · 16 Apr · Shift 1 · Q21

JEE MainChemistrySolutionsMCQ+4 / −1

The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol-1 ) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75%, is :

A
37.5 g
B
75 g
C
150 g
D
50 g

View written solutionFree

Correct answer: NO OPTION MATCHES; CORRECT VALUE SHOULD BE 16.7 G

Use Raoult’s law for a non-volatile solute

For a solution containing a non-volatile solute,

\frac{p}{p^0} = x_{\text{solvent}}

where:

$p^0$ = vapour pressure of pure solvent
$p$ = vapour pressure of solution
$x_{\text{solvent}}$ = mole fraction of solvent

The question says the vapour pressure is reduced to $75\%$ of its original value, so

\frac{p}{p^0} = 0.75

Hence,

x_{\text{solvent}} = 0.75

Find moles of octane

Octane is $\mathrm{C_8H_{18}}$ . Its molar mass is

8(12) + 18(1) = 96 + 18 = 114\ \text{g mol}^{-1}

Given mass of octane = $114\ \text{g}$ , so moles of octane are

n_{\text{solvent}} = \frac{114}{114} = 1\ \text{mol}

Let moles of solute be $n$

Then mole fraction of solvent is

x_{\text{solvent}} = \frac{1}{1+n}

Given this equals $0.75$ :

\frac{1}{1+n} = 0.75 = \frac{3}{4}

So,

4 = 3(1+n)

4 = 3 + 3n

3n = 1

n = \frac{1}{3}\ \text{mol}

Convert moles of solute to mass

Molar mass of solute = $50\ \text{g mol}^{-1}$

\text{mass} = nM = \frac{1}{3} \times 50 = \frac{50}{3} \approx 16.7\ \text{g}

Compare with options

The calculated mass is

\boxed{16.7\ \text{g}}

This does not match any of the given options:

A: $37.5\ \text{g}$
B: $75\ \text{g}$
C: $150\ \text{g}$
D: $50\ \text{g}$

So the data/options appear inconsistent.

Comparison with stored correct answer

Stored correct answer is C (150 g), but the correct calculation gives

\boxed{16.7\ \text{g}}

Therefore, I do not agree with the stored answer.

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