On mixing 3 g of non-volatile solute in 200 mL of water its boiling point (100° C ) becomes 100.52° C. If K b for water is 0.6 K / m then molecular weight of the solute is

Question

On mixing 3 g of non-volatile solute in 200 mL of water its boiling point (100° C ) becomes 100.52° C. If K b for water is 0.6 K / m then molecular weight of the solute is (A) 10.5 g mol -1 (B) 12.6 g mol -1 (C) 15.7 g mol -1 (D) 17.3 g mol -1

Tardigrade Admin · Accepted Answer

Δ T = K b × ( w / m ) × (1000/ W ) 0.52=0.6 × (3/ m ) × (1000/200) m =(1.8 × 5/0.52)=17.3 g mole -1

Q. On mixing $3 g$ of non-volatile solute in $200 m L$ of water its boiling point $(10 0^{\circ} C)$ becomes $100.5 2^{\circ} C$ . If $K_{b}$ for water is $0.6 K / m$ then molecular weight of the solute is

$10.5 g m o l^{- 1}$

$12.6 g m o l^{- 1}$

$15.7 g m o l^{- 1}$

$17.3 g m o l^{- 1}$

$Δ T = K_{b} \times \frac{w}{m} \times \frac{1000}{W}$
$0.52 = 0.6 \times \frac{3}{m} \times \frac{1000}{200}$
$m = \frac{1.8 \times 5}{0.52} = 17.3 g m o l e^{- 1}$

Q. On mixing 3g of non-volatile solute in 200mL of water its boiling point (100∘C) becomes 100.52∘C. If Kb​ for water is 0.6K/m then molecular weight of the solute is

10.5gmol−1

12.6gmol−1

15.7gmol−1

17.3gmol−1