The freezing point depression constant for water is -1.86°C m-1. If 6.00 g Na2SO4 is dissolved in 54.0 g H2O, the freezing point is changed by -3.82°C. Calculate the van’t Hoff factor for Na2SO4.

Question

The freezing point depression constant for water is -1.86°C m-1. If 6.00 g Na2SO4 is dissolved in 54.0 g H2O, the freezing point is changed by -3.82°C. Calculate the van’t Hoff factor for Na2SO4. (A) 2.05 (B) 2.63 (C) 3.11 (D) 0.381

Tardigrade Admin · Accepted Answer

We know that Δ Tf = i × Kf × (wB × 1000/wB × wA) Given: Δ Tf = 3.82, Kf = 1.86 wB = 6 mB = 142, wA = 54 Δ Tf = i× Kf × (wB × 1000/mB × wA) i = (Δ Tf × mB × wA/Kf × wB × 1000) = (3.82 × 142 × 54/1.86 × 6 × 1000) = 2.63

Q. The freezing point depression constant for water is $- 1.8 6^{\circ} C m^{- 1}$ . If $6.00 g N a_{2} S O_{4}$ is dissolved in $54.0 g H_{2} O$ , the freezing point is changed by $- 3.8 2^{\circ} C$ . Calculate the van’t Hoff factor for $N a_{2} S O_{4}$ .

$2.05$

$2.63$

$3.11$

$0.381$

Q. The freezing point depression constant for water is −1.86∘Cm−1. If 6.00gNa2​SO4​ is dissolved in 54.0gH2​O, the freezing point is changed by −3.82∘C. Calculate the van’t Hoff factor for Na2​SO4​.

2.05

2.63

3.11

0.381

We know that ΔTf​=i×Kf​×wB​×wA​wB​×1000​ Given: ΔTf​=3.82,Kf​=1.86 wB​=6mB​=142,wA​=54 ΔTf​=i×Kf​×mB​×wA​wB​×1000​ i=Kf​×wB​×1000ΔTf​×mB​×wA​​ =1.86×6×10003.82×142×54​=2.63

Q. The freezing point depression constant for water is $- 1.8 6^{\circ} C m^{- 1}$ . If $6.00 g N a_{2} S O_{4}$ is dissolved in $54.0 g H_{2} O$ , the freezing point is changed by $- 3.8 2^{\circ} C$ . Calculate the van’t Hoff factor for $N a_{2} S O_{4}$ .

$2.05$

$2.63$

$3.11$

$0.381$