(D) The formula for boiling point elevation is $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.Molality $m = \frac{w_2 \times 10

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(D) The formula for boiling point elevation is $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.Molality $m = \frac{w_2 \times 1000}{M_2 \times w_1}$,where $w_2$ is the mass of solute,$M_2$ is the molar mass,and $w_1$ is the mass of solvent in grams.Substituting the given values: $0.52 = 0.52 \times \frac{12.5 \times 1000}{M_2 \times 250}$.Simplifying the equation: $1 = \frac{12500}{M_2 \times 250}$.$M_2 = \frac{12500}{250} = 50 \ g \ mol^{-1}$.

Class 12 Chemistry Solutions Elevation of boiling point of the solvent

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$A$ solution containing $12.5 \ g$ of non-electrolyte substance in $250 \ g$ of water shows boiling point elevation of $0.52 \ K$. Calculate the molar mass of the substance. $(K_b = 0.52 \ K \ kg \ mol^{-1})$ ......... $g \ mol^{-1}$.

A
$43.06$
B
$25.3$
C
$16.08$
D
$50$

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Elevation of boiling point of the solvent

Calculate $\Delta T_b$ of a $0.45 \text{ m}$ solution of a nonvolatile solute in a solvent if the molal elevation constant of the solvent is $3.0 \text{ K kg mol}^{-1}$. (in $\text{ K}$)

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When $10 \ g$ of a non-volatile solute is dissolved in $100 \ g$ of benzene,it raises the boiling point by $1 \ ^{\circ}C$. The molecular mass of the solute is ....... $g \ mol^{-1}$. ($K_b$ for benzene $= 2.53 \ K \ kg \ mol^{-1}$)

Medium

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When $3 \ g$ of a non-volatile solute is dissolved in $200 \ mL$ of water, the boiling point of the solution becomes $100.52 \ ^oC$. If $K_b$ for water is $0.6 \ K \ kg \ mol^{-1}$, the molar mass of the solute is ......... $g \ mol^{-1}$.

Easy

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$5 \ g$ of a non-volatile organic substance with molar mass $M_A$ is dissolved in $200 \ g$ of tetrahydrofuran. If $K_b$ is the molal elevation constant of tetrahydrofuran,then $\Delta T_b$ will be:

Medium

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The following graph shows the vapour pressure (in $atm$) on the $y$-axis versus temperature $T$ (in $K$) on the $x$-axis for pure water and an aqueous urea solution. What is the boiling point (in $K$) of the urea solution? (Atmospheric pressure $= 1 \ atm$)

MediumAP EAMCET 2024

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$A$ solution containing $12.5 \ g$ of non-electrolyte substance in $250 \ g$ of water shows boiling point elevation of $0.52 \ K$. Calculate the molar mass of the substance. $(K_b = 0.52 \ K \ kg \ mol^{-1})$ ......... $g \ mol^{-1}$.

Similar Questions

Calculate $\Delta T_b$ of a $0.45 \text{ m}$ solution of a nonvolatile solute in a solvent if the molal elevation constant of the solvent is $3.0 \text{ K kg mol}^{-1}$. (in $\text{ K}$)

When $10 \ g$ of a non-volatile solute is dissolved in $100 \ g$ of benzene,it raises the boiling point by $1 \ ^{\circ}C$. The molecular mass of the solute is ....... $g \ mol^{-1}$. ($K_b$ for benzene $= 2.53 \ K \ kg \ mol^{-1}$)

When $3 \ g$ of a non-volatile solute is dissolved in $200 \ mL$ of water, the boiling point of the solution becomes $100.52 \ ^oC$. If $K_b$ for water is $0.6 \ K \ kg \ mol^{-1}$, the molar mass of the solute is ......... $g \ mol^{-1}$.

$5 \ g$ of a non-volatile organic substance with molar mass $M_A$ is dissolved in $200 \ g$ of tetrahydrofuran. If $K_b$ is the molal elevation constant of tetrahydrofuran,then $\Delta T_b$ will be:

The following graph shows the vapour pressure (in $atm$) on the $y$-axis versus temperature $T$ (in $K$) on the $x$-axis for pure water and an aqueous urea solution. What is the boiling point (in $K$) of the urea solution? (Atmospheric pressure $= 1 \ atm$)

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