When 3.00 , g of a substance 'X' is dissolved | vedclass

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(A) The formula for elevation in boiling point is $\Delta T_b = K_b 	imes m$,where $m$ is the molality of the solution.Molality $m = \frac{	ext{mole

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$250$

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(A) The formula for elevation in boiling point is $\Delta T_b = K_b 	imes m$,where $m$ is the molality of the solution.Molality $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{3.00 / M}{100 / 1000} = \frac{30}{M} \, mol \, kg^{-1}$.Given $\Delta T_b = 0.60 \, K$ and $K_b = 5.0 \, K \, kg \, mol^{-1}$.Substituting the values: $0.60 = 5.0 	imes \left( \frac{30}{M} ight)$.$0.60 = \frac{150}{M}$.$M = \frac{150}{0.60} = 250 \, g \, mol^{-1}$.

When $3.00 \, g$ of a substance $'X'$ is dissolved in $100 \, g$ of $CCl_4$,it raises the boiling point by $0.60 \, K$. The molar mass of the substance $'X'$ is $..... \, g \, mol^{-1}$. (Nearest integer).
$[$ Given $K_b$ for $CCl_4$ is $5.0 \, K \, kg \, mol^{-1} ]$

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When $3.00 \, g$ of a substance $'X'$ is dissolved in $100 \, g$ of $CCl_4$,it raises the boiling point by $0.60 \, K$. The molar mass of the substance $'X'$ is $..... \, g \, mol^{-1}$. (Nearest integer).$[$ Given $K_b$ for $CCl_4$ is $5.0 \, K \, kg \, mol^{-1} ]$