The freezing point of a solution composed of 10.0 g of KCl in 100 g of water is 4.5°C. Calculate the van't Hoff factor, i for this solution.

Question

The freezing point of a solution composed of 10.0 g of KCl in 100 g of water is 4.5°C. Calculate the van't Hoff factor, i for this solution. (A) 2.50 (B) 1.8 (C) 1.2 (D) 1.3

Tardigrade Admin · Accepted Answer

Δ Tf = iKf m m = (10× 1000/74.55 ×100) = 1.34 m [Molecular Mass of KCl = 74.55] i = (Δ Tf/Kf m) = (4.5°/(1.86°C/m)(1.34m)) = 1.8

Q. The freezing point of a solution composed of $10.0 g$ of $K Cl$ in $100 g$ of water is $4. 5^{\circ} C$ . Calculate the van't Hoff factor, $i$ for this solution.

2.50

1.8

1.2

1.3

$Δ T_{f} = i K_{f} m$
$m = \frac{10 \times 1000}{74.55 \times 100} = 1.34 m$
[Molecular Mass of $K Cl = 74.55$ ]
$i = \frac{Δ T _{f}}{K _{f} m} = \frac{4. 5 ^{\circ}}{( 1.8 6 ^{\circ} C / m ) ( 1.34 m )} = 1.8$

Q. The freezing point of a solution composed of 10.0g of KCl in 100g of water is 4.5∘C. Calculate the van't Hoff factor, i for this solution.

2.50

1.8

1.2

1.3

ΔTf​=iKf​m m=74.55×10010×1000​=1.34m [Molecular Mass of KCl=74.55] i=Kf​mΔTf​​=(1.86∘C/m)(1.34m)4.5∘​=1.8

Q. The freezing point of a solution composed of $10.0 g$ of $K Cl$ in $100 g$ of water is $4. 5^{\circ} C$ . Calculate the van't Hoff factor, $i$ for this solution.

$Δ T_{f} = i K_{f} m$
$m = \frac{10 \times 1000}{74.55 \times 100} = 1.34 m$
[Molecular Mass of $K Cl = 74.55$ ]
$i = \frac{Δ T _{f}}{K _{f} m} = \frac{4. 5 ^{\circ}}{( 1.8 6 ^{\circ} C / m ) ( 1.34 m )} = 1.8$