A solution of 50 text{ g} of solute 'X' is di | vedclass

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(B) $1$. Elevation in boiling point is given by $\Delta T_b = T_b - T_b^\circ = 357.27 	ext{ K} - 350.06 	ext{ K} = 7.21 	ext{ K}$.$2$. The formula

Vedclass · Accepted Answer

$128$

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(B) $1$. Elevation in boiling point is given by $\Delta T_b = T_b - T_b^\circ = 357.27 	ext{ K} - 350.06 	ext{ K} = 7.21 	ext{ K}$.$2$. The formula for elevation in boiling point is $\Delta T_b = K_b 	imes m$, where $m$ is molality.$3$. Molality $m = \frac{W_2 	imes 1000}{M_2 	imes W_1}$, where $W_2 = 50 	ext{ g}$, $W_1 = 150 	ext{ g}$, and $M_2$ is the molar mass of solute.$4$. Substituting values: $7.21 = 2.77 	imes \frac{50 	imes 1000}{M_2 	imes 150}$.$5$. $M_2 = \frac{2.77 	imes 50000}{7.21 	imes 150} = \frac{138500}{1081.5} \approx 128.06 	ext{ g/mol}$.$6$. Thus, the molar mass is approximately $128 	ext{ g/mol}$.

$A$ solution of $50 \text{ g}$ of solute '$X$' is dissolved in $150 \text{ g}$ of '$Y$' solvent boils at $357.27 \text{ K}$. What is the molar mass of solute (in $\text{ g/mol}$)? (Given $K_b$ and boiling point of pure solvent are $2.77 \text{ K kg/mol}$ and $350.06 \text{ K}$ respectively.)

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