Calculate the molality of a solution of a non-volatile solute if the boiling point of the solution and the molal elevation constant for the solvent are $319.8 \text{ K}$ and $2.5 \text{ K kg mol}^{-1}$ respectively. [Boiling point of pure solvent $= 319.5 \text{ K}$]
- A$0.12 \text{ mol kg}^{-1}$
- B$0.14 \text{ mol kg}^{-1}$
- C$0.17 \text{ mol kg}^{-1}$
- D$0.21 \text{ mol kg}^{-1}$
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Calculate the molality of the solution containing a nonvolatile solute if the boiling point elevation of the solution is $0.39 \ K$.
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[$K_{b}$ of water $= 0.52 \ K \ kg \ mol^{-1}$]
Sea water contains $29.25\%$ $NaCl$ and $19\%$ $MgCl_2$ by weight of solution. The normal boiling point of the sea water is $..........{}^{\circ}C$ (Nearest integer).
Assume $100\%$ ionization for both $NaCl$ and $MgCl_2$.
Given: $K_b(H_2O) = 0.52 \, K \, kg \, mol^{-1}$.
Molar mass of $NaCl$ and $MgCl_2$ is $58.5$ and $95 \, g \, mol^{-1}$ respectively.
Assume $100\%$ ionization for both $NaCl$ and $MgCl_2$.
Given: $K_b(H_2O) = 0.52 \, K \, kg \, mol^{-1}$.
Molar mass of $NaCl$ and $MgCl_2$ is $58.5$ and $95 \, g \, mol^{-1}$ respectively.
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