(C) Given: Degree of association $\alpha = 80\% = 0.8$ and $n = 4$ (for tetramerization). Van't Hoff factor $i = 1 + (\frac{1}{n} - 1)\alpha = 1 + (\f

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(C) Given: Degree of association $\alpha = 80\% = 0.8$ and $n = 4$ (for tetramerization). Van't Hoff factor $i = 1 + (\frac{1}{n} - 1)\alpha = 1 + (\frac{1}{4} - 1) \times 0.8 = 1 - 0.6 = 0.4$. Depression in freezing point $\Delta T_f = i \times K_f \times m$. $0.3 = 0.4 \times 1.86 \times \frac{2.5 \times 1000}{M_A \times 100}$. $0.3 = 0.4 \times 1.86 \times \frac{25}{M_A}$. $M_A = \frac{0.4 \times 1.86 \times 25}{0.3} = \frac{18.6}{0.3} = 62 \text{ g mol}^{-1}$.

Class 12 Chemistry Solutions Abnormal molecular mass

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$A$ non-volatile solute '$A$' tetramerizes in water to the extent of $80\%$. $2.5 \text{ g}$ of '$A$' in $100 \text{ g}$ of water lowers the freezing point by $0.3^\circ \text{C}$. The molar mass of $A$ in $\text{g mol}^{-1}$ is ($K_f$ for water $= 1.86 \text{ K kg mol}^{-1}$)

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$0.2$ molal aqueous solution of $KCl$ freezes at $-0.680^{\circ} C$. Calculate van't Hoff factor for this solution. $(K_{f} = 1.86 \ K \ kg \ mol^{-1})$

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Calculate the van't Hoff factor $(i)$ for an aqueous solution of $0.02 \text{ m}$ formic acid if it freezes at $-0.045^{\circ}\text{C}$. Given: $K_f = 1.86 \text{ K kg mol}^{-1}$ and the freezing point of pure water $= 0^{\circ}\text{C}$.

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$A$ non-volatile solute '$A$' tetramerizes in water to the extent of $80\%$. $2.5 \text{ g}$ of '$A$' in $100 \text{ g}$ of water lowers the freezing point by $0.3^\circ \text{C}$. The molar mass of $A$ in $\text{g mol}^{-1}$ is ($K_f$ for water $= 1.86 \text{ K kg mol}^{-1}$)

Similar Questions

In a solution of sulfur in carbon disulfide,$80 \%$ of the sulfur exists as $S_8$ and the remaining exists as $S_2$. The van't Hoff factor '$i$' will be:

Which of the following expressions correctly defines the Van't Hoff factor $i$?

The molecular weight of $NaCl$ determined by osmotic pressure method will be

$0.2$ molal aqueous solution of $KCl$ freezes at $-0.680^{\circ} C$. Calculate van't Hoff factor for this solution. $(K_{f} = 1.86 \ K \ kg \ mol^{-1})$

Calculate the van't Hoff factor $(i)$ for an aqueous solution of $0.02 \text{ m}$ formic acid if it freezes at $-0.045^{\circ}\text{C}$. Given: $K_f = 1.86 \text{ K kg mol}^{-1}$ and the freezing point of pure water $= 0^{\circ}\text{C}$.

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