The molal boiling point constant for water is | vedclass

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Question

(C) The molality $(m)$ of the solution is calculated as: $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.1 \, mol}{0.2 \, k

Vedclass · Accepted Answer

$100.25$

Vedclass · Answer

(C) The molality $(m)$ of the solution is calculated as: $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.1 \, mol}{0.2 \, kg} = 0.5 \, mol \, kg^{-1}$.Using the formula for elevation in boiling point: $\Delta T_b = K_b 	imes m$.$\Delta T_b = 0.513 \, ^oC \, kg \, mol^{-1} 	imes 0.5 \, mol \, kg^{-1} = 0.2565 \, ^oC$.The boiling point of the solution $(T_b)$ is: $T_b = T_b^0 + \Delta T_b = 100 \, ^oC + 0.2565 \, ^oC = 100.2565 \, ^oC$.Rounding to two decimal places,the value is approximately $100.25 \, ^oC$.

The molal boiling point constant for water is $0.513 \, ^oC \, kg \, mol^{-1}$. When $0.1 \, mol$ of sugar is dissolved in $200 \, mL$ of water,the solution boils under a pressure of one atmosphere at ......... $^oC$.

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