Question

At ${100}^{\circ }C$ the vapour pressure of a solution of $6.5g$ of a solute in $100g$ water in $732mm$. If $K_b=0.52$, the boiling point of this solution will be

• A. ${100}^{\circ }C$
• B. ${102}^{\circ }C$
• C. ${103}^{\circ }C$
• D. ${101}^{\circ }C$

Answer 1

The correct option is D ${101}^{\circ }C$

Elevation in boiling point of a solvent upon addition of solute is given as:

$\Delta T_b=m.K_b$

m=molality of the solution=number of moles of solute in 1 kg solvent

$K_b$=ebullioscopic constant$={0.52}^{0}kg/mol$

Vapour pressure of solution $P_s=732mmofHg$

Vapour pressure of pure solvent (at ${100}^{0}C$) P=760 mm of Hg

mass of solute, $m_s=6.5g$, mass of water, $m_w=100g$, molar mass of water, $M_w=18g/mol$

let molar mass of solute be $M_s$

thus using $\frac{P-P_s}{P_s}=\frac{n_s}{n_w}$

$\therefore \frac{760-732}{732}=\frac{m_s\times M_w}{M_s\times m_w}$

or $\frac{28}{732}=\frac{6.5\times 18}{M_s\times 100}$

or $M_s=30.6g/mol$

thus $\Delta T_b=0.52\times \frac{m_s\times 1000}{M_s\times m_w}$

$\Delta T_b=0.52\times \frac{6.5\times 1000}{30.6\times 100}={1.105}^{0}C$

thus boiling point of water is

$T_{solvent}=100+1.105={101.105}^{0}C$