At certain Hill-station pure water boils at 99.72 ° C . If Kb for water is 0.513 ° Ckgmol- 1 . The boiling point of 0.69 m solution of urea will be-

Question

At certain Hill-station pure water boils at 99.72 ° C . If Kb for water is 0.513 ° Ckgmol- 1 . The boiling point of 0.69 m solution of urea will be- (A) 100.074 ° C (B) 103 ° C (C) 100.359 ° C (D) Un predictable

Tardigrade Admin · Accepted Answer

Δ Tb=Kb× m =0.513× 0.69 ∴ Δ Tb=0.354 So, Tb=0.354+99.72=100.074 ° C

Q. At certain Hill-station pure water boils at $99.72$ $^{\circ} C$ . If $K_{b}$ for water is $0.513$ $^{\circ} C k g m o l^{- 1}$ . The boiling point of $0.69 m$ solution of urea will be-

$100.074$ $^{\circ} C$

$103$ $^{\circ} C$

$100.359$ $^{\circ} C$

Un predictable

$Δ T_{b} = K_{b} \times m$
$= 0.513 \times 0.69$
$∴ Δ T_{b} = 0.354$
So, $T_{b} = 0.354 + 99.72 = 100.074$ $^{\circ} C$

Q. At certain Hill-station pure water boils at 99.72 ∘C . If Kb​ for water is 0.513 ∘Ckgmol−1 . The boiling point of 0.69m solution of urea will be-

100.074 ∘C

103 ∘C

100.359 ∘C