Question

For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is $2^{o}C$. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is:

[$K_b=0.76KKgmo{l}^{-1}$]

• A. $724$
• B. $740$
• C. $736$
• D. $718$

Answer 1

Answer: (A)

$724$

The vapour pressure of the solution is 724 mm of Hg.
Given : $w_{solute}=2.5g,W_{solvent}=100g$

$\Delta T_b=2^o$

$\therefore \Delta T_b=\frac{1000\times K_b\times w}{W\times m}$

or $2=\frac{1000\times 0.76\times 2.5}{100\times m}$

$\therefore m=9.5$

Now, $\frac{P^0-P_S}{P^0}=\frac{w\times m}{m\times W}$ (Given dilute solution)

$\frac{760-P_S}{760}=\frac{2.5\times 18}{9.5\times 100}=0.047$

$\therefore P_S=724mmofHg$