(A) The elevation in boiling point is given by the formula: $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution. Molality $m$ is def

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$M_2 = \frac{K_b \times W_2 \times 1000}{\Delta T_b \times W_1}$

(A) The elevation in boiling point is given by the formula: $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution. Molality $m$ is defined as the number of moles of solute $(n_2)$ per kilogram of solvent ($W_1$ in grams): $m = \frac{W_2 \times 1000}{M_2 \times W_1}$. Substituting this into the elevation formula: $\Delta T_b = \frac{K_b \times W_2 \times 1000}{M_2 \times W_1}$. Rearranging to solve for the molar mass of the solute $(M_2)$: $M_2 = \frac{K_b \times W_2 \times 1000}{\Delta T_b \times W_1}$.

Class 12 Chemistry Solutions Elevation of boiling point of the solvent

Medium

The correct relation between the elevation in boiling point and the molar mass of a solute is: ($W_1$ and $W_2$ are respectively weight of solvent $\&$ solute)

MHT CET 2018 All MHT CET

A
$M_2 = \frac{K_b \times W_2 \times 1000}{\Delta T_b \times W_1}$
B
$M_2 = \frac{K_b \times W_1 \times 1000}{\Delta T_b \times W_2}$
C
$M_2 = \frac{\Delta T_b \times K_b \times 1000}{W_1 \times W_2}$
D
$M_2 = \frac{\Delta T_b \times W_1 \times 1000}{K_b \times W_2}$

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Elevation of boiling point of the solvent

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The boiling point of water $(100\,^{\circ}C)$ becomes $100.52\,^{\circ}C$,if $3\,g$ of a non-volatile solute is dissolved in $200\,mL$ of water. The molecular weight of the solute is: (${K_b}$ for water is $0.6\,K\,kg\,mol^{-1}$)

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What is the boiling point of a solution of $0.1 \ m$ $KCl$ (in $K$)? $K_{b}$ of water is $0.52 \ K \ kg \ mol^{-1}$ and $\alpha = 100 \ \%$ (boiling point of pure water is $373 \ K$).

MediumAP EAMCET 2023

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$A$ solution is prepared by dissolving a $2.5 \ g$ sample of an unknown compound in $34 \ g$ of benzene,$C_6H_6$. The solution boils at $1.38 \ ^\circ C$ higher than pure benzene. Which expression gives the molar mass of the unknown compound? $K_b$ of $C_6H_6$ is $2.53 \ ^\circ C \ m^{-1}$.

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Explain why on addition of $1 \ mol$ of $NaCl$ to $1 \ litre$ of water,the boiling point of water increases,while addition of $1 \ mol$ of methyl alcohol to one litre of water decreases its boiling point.

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$Y \ g$ of non-volatile organic substance of molecular mass $M$ is dissolved in $250 \ g$ benzene. The molal elevation constant of benzene is $K_b$. The elevation in its boiling point is given by:

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The correct relation between the elevation in boiling point and the molar mass of a solute is: ($W_1$ and $W_2$ are respectively weight of solvent $\&$ solute)

Similar Questions

The boiling point of water $(100\,^{\circ}C)$ becomes $100.52\,^{\circ}C$,if $3\,g$ of a non-volatile solute is dissolved in $200\,mL$ of water. The molecular weight of the solute is: (${K_b}$ for water is $0.6\,K\,kg\,mol^{-1}$)

What is the boiling point of a solution of $0.1 \ m$ $KCl$ (in $K$)? $K_{b}$ of water is $0.52 \ K \ kg \ mol^{-1}$ and $\alpha = 100 \ \%$ (boiling point of pure water is $373 \ K$).

$A$ solution is prepared by dissolving a $2.5 \ g$ sample of an unknown compound in $34 \ g$ of benzene,$C_6H_6$. The solution boils at $1.38 \ ^\circ C$ higher than pure benzene. Which expression gives the molar mass of the unknown compound? $K_b$ of $C_6H_6$ is $2.53 \ ^\circ C \ m^{-1}$.

Explain why on addition of $1 \ mol$ of $NaCl$ to $1 \ litre$ of water,the boiling point of water increases,while addition of $1 \ mol$ of methyl alcohol to one litre of water decreases its boiling point.

$Y \ g$ of non-volatile organic substance of molecular mass $M$ is dissolved in $250 \ g$ benzene. The molal elevation constant of benzene is $K_b$. The elevation in its boiling point is given by:

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