To 500 cm 3 of water, 3 × 10-3 kg of CH 3 COOH is added. If 23 % of CH 3 COOH is disociated, what will be the depression in freezing point- (Kf. for water is 1.86 k kg mol -1 )

Question

To 500 cm 3 of water, 3 × 10-3 kg of CH 3 COOH is added. If 23 % of CH 3 COOH is disociated, what will be the depression in freezing point- (Kf. for water is 1.86 k kg mol -1 ) (A) 0.372 K (B) 0.228 K (C) 0.328 K (D) 0.556 K

Tardigrade Admin · Accepted Answer

Apply Δ T f = i K f m Δ T f = i × 1.86 Δ T f = i × 1.86 × ((3/60)/(500)1000) ldots ldots(1) (ρ of H 2 O =1 g / cm 3) So, 500 cm 3 water will weigh 500 g Also, α=(i-1/n-1) ⇒ 0.23=(i-1/2-1) i=1.23 Placing the value of (i) in equation 1, we get Δ T f =1.23 × 1.86 × (3/60) × (1000/500) Δ T f =0.228 K

Q. To 500cm3 of water, 3×10−3kg of CH3​COOH is added. If 23% of CH3​COOH is disociated, what will be the depression in freezing point- (Kf​ for water is 1.86kkg mol−1 )

0.372 K

0.228 K

0.328 K

0.556 K

Apply ΔTf​=iKf​m ΔTf​=i×1.86 ΔTf​=i×1.86×1000500​603​​……(1) (ρ of H2​O=1g/cm3) So, 500cm3 water will weigh 500g Also, α=n−1i−1​ ⇒0.23=2−1i−1​ i=1.23 Placing the value of (i) in equation 1, we get ΔTf​=1.23×1.86×603​×5001000​ ΔTf​=0.228K

Q. To $500 c m^{3}$ of water, $3 \times 1 0^{- 3} k g$ of $C H_{3} COO H$ is added. If $23%$ of $C H_{3} COO H$ is disociated, what will be the depression in freezing point- $(K_{f}$ for water is $1.86 k k g$ $m o l^{- 1}$ )