Calculate the approximate Δ Tb (in .K ) for 0.001 molal KCl solution, if its van't-Hoff factor is 1.98[Kb. of water is .0.52 K kg mol -1]

Question

Calculate the approximate Δ Tb (in .K ) for 0.001 molal KCl solution, if its van't-Hoff factor is 1.98[Kb. of water is .0.52 K kg mol -1] (A) 1.03 (B) 1.03 × 10-3 (C) 1.03 × 10-5 (D) 1.03 × 10-1

Tardigrade Admin · Accepted Answer

We know that Δ Tb=i Kb m (∵ Δ Tb=i . Kb × m) where, Δ Tb= elivation in boiling point (to time) i = van' t - Hoff factor (1.98) Kb = molal elevation constant (0.52) m = molality (0.001 m ) Δ Tb =1.98 × 0.52 × 0.001=0.0010296 Δ Th =1.03 × 10-3

Q. Calculate the approximate $Δ T_{b}$ (in $K)$ for $0.001$ molal $K Cl$ solution, if its van't-Hoff factor is $1.98 [K_{b}$ of water is $0.52 K k g m o l^{- 1}]$

$1.03$

$1.03 \times 1 0^{- 3}$

$1.03 \times 1 0^{- 5}$

$1.03 \times 1 0^{- 1}$

Q. Calculate the approximate ΔTb​ (in K) for 0.001 molal KCl solution, if its van't-Hoff factor is 1.98[Kb​ of water is 0.52Kkgmol−1]

1.03

1.03×10−3

1.03×10−5

1.03×10−1

We know that ΔTb​=iKb​m (∵ΔTb​=i.Kb​×m) where, ΔTb​= elivation in boiling point (to time) i = van' t - Hoff factor (1.98) Kb​ = molal elevation constant (0.52) m = molality (0.001m) ΔTb​=1.98×0.52×0.001=0.0010296 ΔTh​=1.03×10−3

Q. Calculate the approximate $Δ T_{b}$ (in $K)$ for $0.001$ molal $K Cl$ solution, if its van't-Hoff factor is $1.98 [K_{b}$ of water is $0.52 K k g m o l^{- 1}]$

$1.03$

$1.03 \times 1 0^{- 3}$

$1.03 \times 1 0^{- 5}$

$1.03 \times 1 0^{- 1}$