Question

The relative lowering of the vapour pressure of an aqueous solution containing a non-volatile solute is $0.0125$. The molality of the solution is:

• A. $0.07$
• B. $0.30$
• C. $0.70$
• D. $0.125$

Answer 1

The correct option is C

$0.70$

The explanation for the correct option:

(c) $0.70$

• The mole fraction of the solute is equal to the relative decrease in vapour pressure.

Step 1: Applying Raoult's law:

• ${\mathrm{X}}_{\mathrm{B}}=\frac{\mathrm{m}.{\mathrm{M}}_{\mathrm{A}}}{1000+\mathrm{m}.{\mathrm{M}}_{\mathrm{A}}}$

Step 2: Analyzing the given quantities

• ${\mathrm{X}}_{\mathrm{B}}=$Relative lowering of vapour pressure
• ${\mathrm{M}}_{\mathrm{A}}=$ Molecular mass of ${\mathrm{H}}_2\mathrm{O}$
• $\mathrm{m}=$molality
• ${\mathrm{X}}_{\mathrm{B}}=$$0.0125$
• ${\mathrm{M}}_{\mathrm{A}}=$$16+1\times 2=18$

Step 3: Finding molality

• $$\begin{gathered} 0.0125=\frac{\mathrm{m}\times 18}{1000+\mathrm{m}\times 18} \\ 12.5+0.225\mathrm{m}=18\mathrm{m} \\ 17.775\mathrm{m}=12.5 \\ \mathrm{m}=\frac{12.5}{17.775} \\ \mathrm{m}=0.70 \end{gathered}$$
• Thus molality of the solution is $0.70\mathrm{m}$.
• Hence this option is correct

The explanation for the incorrect options:

Since the molality is $0.70\mathrm{m}$. Therefore, options (a), (b) and (d) are incorrect.

Hence the correct option is (c ) $0.70$.