A solution of 5.6 g nonvolatile solute in 50 | 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.Molality $m = \frac{W_2 	imes 1000}{M_2 \

Vedclass · Accepted Answer

$192 \ g \ mol^{-1}$

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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.Molality $m = \frac{W_2 	imes 1000}{M_2 	imes W_1}$,where $W_2$ is the mass of solute,$M_2$ is the molar mass of solute,and $W_1$ is the mass of solvent.Substituting the values: $1.75 = \frac{3 	imes 5.6 	imes 1000}{M_2 	imes 50}$.Rearranging for $M_2$: $M_2 = \frac{3 	imes 5.6 	imes 1000}{1.75 	imes 50}$.$M_2 = \frac{16800}{87.5} = 192 \ g \ mol^{-1}$.

$A$ solution of $5.6 \ g$ nonvolatile solute in $50 \ g$ solvent has an elevation in boiling point of $1.75 \ K$. What is the molar mass of the solute $\left(K_{b} = 3 \ K \ kg \ mol^{-1}\right)$?

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