Question

Molal elevation constant for benzene is 2.52K/m . A solution of some organic substance in benzene boils at 0.126°C higher than benzene. what is molality of the solution .​

Answer 1

Answer:

∆Tb = Kb ×m

m = molality

m = ∆Tb/Kb

m = .126/2.52

m = 0.05

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Answer 2

Final answer: The molality of the solution $= 0.05$ $\frac{mol}{kg}$

Given that: We are given a solution of some organic substance in benzene boils at $0.126\°C$ higher than benzene. The molal elevation constant for benzene is $2.52$ $\frac{K}{m}$.

To find: We have to find the molality of the solution.​

Explanation:

• Elevation in boiling point ($\triangle T_{b}$) is directly proportional to the molality ($m$) of the solution.

$\triangle T_{b}$ $\alpha$ $m$

$\triangle T_{b}=K_{b} \times m$

$K_{b}$ = molal elevation constant.

• The molal elevation constant of a solvent is defined as the elevation in boiling point caused by dissolving one mole of a non-volatile solute in $1000$ $g$ of solvent.
• The unit of $K_{b}$ is $K kg mol^{-1}$ or it may be represented by $\frac{K}{m}$.

$Molal elevation constant(K_{b})=\frac{ R(T_{b}^{0})^{2}M_{1}}{\triangle _{vap}H}$

Where $R$ is the universal gas constant, and $T_{b}^{0}$, $M_{1}$, $\triangle _{vap}H$ are respectively the boiling point, molar mass, and molar heat of vaporization of the solvent in SI units.

• Here, $K_{b}=2.52\frac{K}{m}$

$\triangle T_{b}$ $= 0.126\°C$ is taken as $0.126$ $K$

$m=?$

$m = \frac{\triangle T_{b}}{K_{b}}$

• Substitute the values of $\triangle T_{b}$ and $K_{b}$.

$m=\frac{0.126K}{2.52(\frac{K}{m})}=0.05m=0.05 \frac{mol}{kg}$

• Hence, the molality of the solution $= 0.05$ $\frac{mol}{kg}$

To know more about the concept please go through the links

https://brainly.in/question/19984766

https://brainly.in/question/18769668

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