The freezing point depression constant for wa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The formula for freezing point depression is $\Delta T_f = i 	imes K_f 	imes m.$ Here,$\Delta T_f = 3.82 \, ^oC,$ $K_f = 1.86 \, ^oC \, kg \, mo

Vedclass · Accepted Answer

$2.63$

Vedclass · Answer

(B) The formula for freezing point depression is $\Delta T_f = i 	imes K_f 	imes m.$ Here,$\Delta T_f = 3.82 \, ^oC,$ $K_f = 1.86 \, ^oC \, kg \, mol^{-1},$ $W_{solute} = 5.00 \, g,$ $W_{solvent} = 45.0 \, g,$ and the molar mass of $Na_2SO_4$ is $M = 142 \, g \, mol^{-1}.$ The molality $m$ is given by $\frac{W_{solute} 	imes 1000}{M 	imes W_{solvent}} = \frac{5.00 	imes 1000}{142 	imes 45.0} \approx 0.782 \, m.$ Substituting these values into the equation: $3.82 = i 	imes 1.86 	imes 0.782.$ Solving for $i$: $i = \frac{3.82}{1.86 	imes 0.782} \approx 2.63.$

The freezing point depression constant for water is $1.86 \, ^oC \, kg \, mol^{-1}.$ If $5.00 \, g$ of $Na_2SO_4$ is dissolved in $45.0 \, g$ of $H_2O,$ the freezing point is lowered by $3.82 \, ^oC.$ Calculate the van't Hoff factor for $Na_2SO_4.$

Similar Questions

At higher altitude,the boiling point of water is $95\,^oC$. The amount of $NaCl$ added to $1\, kg$ of water $(K_b = 0.52\, K\, kg\, mol^{-1})$ in order to raise the boiling point of the solution to $100\,^oC$ is ......... $g$. (Assuming $90\%$ ionization of $NaCl$)

The relationship between the van’t Hoff factor $(i)$ and the degree of dissociation $(\alpha)$ is:

$K_2HgI_4$ is $40 \%$ ionized in aqueous solution. The value of its van't Hoff factor $(i)$ is:

Of the following $0.10 \ m$ aqueous solutions,which one will exhibit the largest freezing point depression?

If $13.44 \ g$ of $CuCl_2$ is dissolved in $1 \ kg$ of water,calculate the elevation in boiling point of the solution. $[K_b = 0.5 \ K \ kg \ mol^{-1}$,molar mass of $CuCl_2 = 134.1 \ g \ mol^{-1}]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module