Question

The percentage composition by weight of an aqueous solution of a solute (molar mass 150) which boils at 373.26 is:

$(Given:K_b$ = 0.52)

• A. 5
• B. 15
• C. 7
• D. 10

Answer 1

Answer: (C)

7

$\Delta T_b=K_b\times m$

$\Delta T_b$ $=$Elevation in boiling point

$K_b$ $=$ Ebulloiscopic constant$=0.52$

$\Delta T_b=0.26$

$m=$ Molality of the solution

$m=\frac{\Delta T_b}{K_b}=\frac{0.26}{0.52}=0.5$

Molality is defined as the number of moles of solution in $1000$ g of solvent

$1000g$ of water will contain moles of solute=$0.5$ mole

Therefore $0.5\times 150=75g$ of solute in $1000g$ or $1kg$ of the water

$\Delta T_b=\frac{K_b\times w_2\times 1000}{w_1\times M_1}0.26=\frac{0.52\times w_2\times 1000}{1000\times 150}{w}_2=75$

Percentage composition$=\frac{75\times 100}{1000}=7.5$%