The increase in boiling point of a solution containing 0.6 g urea in 200 gram water is 0.50°C.Find the molal elevation constant

Question

The increase in boiling point of a solution containing 0.6 g urea in 200 gram water is 0.50°C.Find the molal elevation constant (A) 10 K gram mol-1 (B) 1.0 K mol (C) 10 K kg mol-1 (D) 10 K kg mol

Tardigrade Admin · Accepted Answer

Molality, m = (0.6/60) × (1000/200) = 0.05 m Δ Tb = 0.5 K now Δ Tb = Kb × m or Kb = (Δ Tb/m) = (0.5/0.05) = 10 K kg mol-1

Q. The increase in boiling point of a solution containing 0.6 g urea in 200 gram water is 0.50 $^{\circ}$ C.Find the molal elevation constant

10 K gram $m o l^{- 1}$

1.0 K $m o l$

10 K kg $m o l^{- 1}$

10 K kg mol

Molality, $m = \frac{0.6}{60} \times \frac{1000}{200}$ = 0.05 m
$Δ T_{b}$ = 0.5 K
now $Δ T_{b} = K_{b} \times m$
or $K_{b} = \frac{Δ T _{b}}{m} = \frac{0.5}{0.05}$ = 10 K kg mol $^{- 1}$

Q. The increase in boiling point of a solution containing 0.6 g urea in 200 gram water is 0.50∘C.Find the molal elevation constant

10 K gram mol−1

1.0 K mol

10 K kg mol−1