What is elevation in boiling point? Explain.
(N/A) The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure $(1.013 \ bar)$.
When a non-volatile solute is added to a pure solvent,the vapour pressure of the resulting solution decreases compared to the pure solvent at the same temperature.
Because the vapour pressure of the solution is lower,it must be heated to a higher temperature to reach the atmospheric pressure $(1.013 \ bar)$.
This increase in the boiling point of the solution compared to the pure solvent is known as the elevation in boiling point,denoted by $\Delta T_b$.
When a non-volatile solute is added to a pure solvent,the vapour pressure of the resulting solution decreases compared to the pure solvent at the same temperature.
Because the vapour pressure of the solution is lower,it must be heated to a higher temperature to reach the atmospheric pressure $(1.013 \ bar)$.
This increase in the boiling point of the solution compared to the pure solvent is known as the elevation in boiling point,denoted by $\Delta T_b$.
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View Solution$A$ solution of non-volatile solute has boiling point elevation $0.5 \ K$. Calculate molality of solution $[K_{b} = 2.40 \ K \ kg \ mol^{-1}]$.
An aqueous dilute solution containing non-volatile solute boils at $100.52^{\circ} C$. What is the molality of the solution (in $m$)? ($K_b = 0.52 \ K \ kg \ mol^{-1}$, boiling temperature of water $= 100^{\circ} C$)
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.
DifficultJEE MAIN 2023
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