(C) Molar mass of $K_2SO_4 = 2 \times 39 + 32 + 4 \times 16 = 174 \ g/mol$. Molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$. Molarity of $K_2SO_4$ $(C

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(C) Molar mass of $K_2SO_4 = 2 \times 39 + 32 + 4 \times 16 = 174 \ g/mol$. Molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$. Molarity of $K_2SO_4$ $(C_1) = \frac{17.4 \times 1000}{174 \times 100} = 1 \ M$. Molarity of $NaCl$ $(C_2) = \frac{5.85 \times 1000}{58.5 \times 100} = 1 \ M$. For isotonic solutions,$i_1 C_1 = i_2 C_2$. For $NaCl$,$i_2 = 2$ (as it is $100\%$ ionized). $i_1 \times 1 = 2 \times 1 \Rightarrow i_1 = 2$. For $K_2SO_4$,$i_1 = 1 + (n-1)\alpha = 1 + (3-1)\alpha = 1 + 2\alpha$. $1 + 2\alpha = 2$ $\Rightarrow 2\alpha = 1$ $\Rightarrow \alpha = 0.5$. $\%$ ionization $= 0.5 \times 100 = 50\%$.

Class 12 Chemistry Solutions Colligative properties of electrolyte

Difficult

$17.4\% \text{ (mass/volume) } K_2SO_4$ solution at $27^\circ C$ is isotonic to $5.85\% \text{ (mass/volume) } NaCl$ solution at $27^\circ C$. If $NaCl$ is $100\%$ ionized,what is the $\%$ ionization of $K_2SO_4$ in aqueous solution? [Atomic weights: $K = 39, Na = 23, S = 32, O = 16, Cl = 35.5$]

A
$25$
B
$75$
C
$50$
D
\text{None of these}

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Colligative properties of electrolyte

The measured osmotic pressure of a solution prepared by dissolving $17.4 \ mg$ of $K_2SO_4$ in $2 \ L$ of water at $27^{\circ} C$ is $3.735 \times 10^{-3} \ bar$. The van't Hoff factor is $(R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$; atomic weights $K = 39, S = 32, O = 16)$.

MediumAP EAMCET 2018

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Of the following four aqueous solutions,the total number of those solutions whose freezing point is lower than that of $0.10 \, M \, C_{2}H_{5}OH$ is (Integer answer).
$(i)$ $0.10 \, M \, Ba_{3}(PO_{4})_{2}$
$(ii)$ $0.10 \, M \, Na_{2}SO_{4}$
$(iii)$ $0.10 \, M \, KCl$
$(iv)$ $0.10 \, M \, Li_{3}PO_{4}$

MediumJEE MAIN 2021

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Which of the following will show maximum osmotic pressure for its $1 \, m$ solution?

Medium

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Calculate the molality of an aqueous solution of an electrolyte that freezes at $-0.93 \text{ }^\circ\text{C}$. Given that $K_f$ for water is $1.86 \text{ K kg mol}^{-1}$ and the van't Hoff factor $(i)$ is $1.25$. (Freezing point of pure water $= 0 \text{ }^\circ\text{C}$) (in $\text{ m}$)

DifficultMHT CET 2026

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Molal depression constant for a solvent is $4.0 \, K \, kg \, mol^{-1}$. The depression in the freezing point of the solvent for $0.03 \, mol \, kg^{-1}$ solution of $K_2SO_4$ is .............. $K$ (Assume complete dissociation of the electrolyte).

DifficultJEE MAIN 2019

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$17.4\% \text{ (mass/volume) } K_2SO_4$ solution at $27^\circ C$ is isotonic to $5.85\% \text{ (mass/volume) } NaCl$ solution at $27^\circ C$. If $NaCl$ is $100\%$ ionized,what is the $\%$ ionization of $K_2SO_4$ in aqueous solution? [Atomic weights: $K = 39, Na = 23, S = 32, O = 16, Cl = 35.5$]

Similar Questions

The measured osmotic pressure of a solution prepared by dissolving $17.4 \ mg$ of $K_2SO_4$ in $2 \ L$ of water at $27^{\circ} C$ is $3.735 \times 10^{-3} \ bar$. The van't Hoff factor is $(R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$; atomic weights $K = 39, S = 32, O = 16)$.

Which of the following will show maximum osmotic pressure for its $1 \, m$ solution?

Calculate the molality of an aqueous solution of an electrolyte that freezes at $-0.93 \text{ }^\circ\text{C}$. Given that $K_f$ for water is $1.86 \text{ K kg mol}^{-1}$ and the van't Hoff factor $(i)$ is $1.25$. (Freezing point of pure water $= 0 \text{ }^\circ\text{C}$) (in $\text{ m}$)

Molal depression constant for a solvent is $4.0 \, K \, kg \, mol^{-1}$. The depression in the freezing point of the solvent for $0.03 \, mol \, kg^{-1}$ solution of $K_2SO_4$ is .............. $K$ (Assume complete dissociation of the electrolyte).

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