How many grams of KCl should be added to 0.1 kg of water to lower its freezing point to -8.0° C ?[Kf. for water =1.86° C kg mol -1 )

Question

How many grams of KCl should be added to 0.1 kg of water to lower its freezing point to -8.0° C ?[Kf. for water =1.86° C kg mol -1 ) (A) 160.2 (B) 150.2 (C) 140.2 (D) 130.2

Tardigrade Admin · Accepted Answer

Δ Tf=(Kf × w × 1000/m × W) Given, Δ Tf=8o C Kf=1.86° C kg mol -1 m= mass of solute =35.5+19=54.5 w =1 kg ∴ 8=(1.86 × w × 100/54.5 × 1) W =320.5 g KClK++C l-(i=2) Hence, w=(w/2)=(320.5/2)=160.2 g

Q. How many grams of KCl should be added to 0.1kg of water to lower its freezing point to −8.0∘C?[Kf​ for water =1.86∘Ckgmol−1 )

160.2

150.2

140.2

130.2

ΔTf​=m×WKf​×w×1000​ Given, ΔTf​=8oCKf​=1.86∘Ckgmol−1 m= mass of solute =35.5+19=54.5 w=1kg ∴8=54.5×11.86×w×100​W=320.5g KClK++Cl−(i=2) Hence, w=2w​=2320.5​=160.2g

Q. How many grams of $K Cl$ should be added to $0.1 k g$ of water to lower its freezing point to $- 8. 0^{\circ} C ? [K_{f}$ for water $= 1.8 6^{\circ} C k g m o l^{- 1}$ )