Question

Lowering of vapour pressure of $1.00m$ aqueous solution of a non volatile solute in hypothetical solvent of molar mass $40g$ at its normal boiling point is:

• A. $29.24torr$
• B. $30.4torr$
• C. $35.00torr$
• D. $40.00torr$

Answer 1

Answer: (B)

$30.4torr$

According to Raoult's law of relative lowering of vapour pressure upon addition of solute to a solvent is given by:

$\frac{\Delta p}{p_{solution}}=\frac{m\times M_{solvent}}{1000}$

where $\Delta p$=lowering of vapour pressure,

$p_{solution}$=pressure of solution=normal boiling point = $1atm=760torr$

$m=$molality$=1\ m$,

$M_{ solvent }=$molecular mass of solvent.

Upon substitution we get:

$\frac{\Delta p}{760}=\frac{1\times 40}{1000}$

$\Delta p=30.4torr$

Hence, the correct option is $\text{B}$