The boiling point (in ° C ) of 0.1 molal aqueous solution of CuSO 4 .5 H 2 O at 1 bar is closest to: [Given: Ebullioscopic (molal boiling point elevation) constant of water, .K b =0.512 K Kg mol -1]:-

Question

The boiling point (in ° C ) of 0.1 molal aqueous solution of CuSO 4 .5 H 2 O at 1 bar is closest to: [Given: Ebullioscopic (molal boiling point elevation) constant of water, .K b =0.512 K Kg mol -1]:- (A) 100.36 (B) 99.64 (C) 100.10 (D) 99.90

Tardigrade Admin · Accepted Answer

CuSO4 ⋅ 5H2O ->[H2O] Cu2⊕ + SO42 ominus i = 2 Δ Tb = i . Kb . m = 2 × 0.512 × 0.1 = 0.1024 Tb' = Tb 0 + Δ Tb = 100 + 0.1024 = 100.10

Q. The boiling point (in $^{\circ} C$ ) of $0.1$ molal aqueous solution of $C u S O_{4} .5 H_{2} O$ at $1$ bar is closest to: [Given: Ebullioscopic (molal boiling point elevation) constant of water, $K_{b} = 0.512 K K g m o l^{- 1}] : -$

100.36

99.64

100.10

99.90

$C u S O_{4} \cdot 5 H_{2} O H X_{2} O C u^{2 \oplus} + S O_{4}^{2 ⊖}$
$i = 2$
$Δ T_{b} = i . K_{b} . m = 2 \times 0.512 \times 0.1 = 0.1024$
$T_{b}^{'} = T_{b}^{0} + Δ T_{b} = 100 + 0.1024 = 100.10$

Q. The boiling point (in ∘C ) of 0.1 molal aqueous solution of CuSO4​.5H2​O at 1 bar is closest to: [Given: Ebullioscopic (molal boiling point elevation) constant of water, Kb​=0.512KKgmol−1]:−