Bromoform has a normal freezing point of 7.73 | vedclass

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(B) The depression in freezing point is given by $\Delta T_{f} = T_{f}^{\circ} - T_{f} = 7.734^{\circ} C - 5.43^{\circ} C = 2.304^{\circ} C$. The form

Vedclass · Accepted Answer

$162.5$

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(B) The depression in freezing point is given by $\Delta T_{f} = T_{f}^{\circ} - T_{f} = 7.734^{\circ} C - 5.43^{\circ} C = 2.304^{\circ} C$. The formula for depression in freezing point is $\Delta T_{f} = K_{f} 	imes m$,where $m$ is the molality. Molality $m = \frac{w_{solute} 	imes 1000}{M_{solute} 	imes w_{solvent} 	ext{ (in g)}}$. Substituting the values: $2.304 = 14.4 	imes \frac{2.60 	imes 1000}{M_{solute} 	imes 100}$. $2.304 = 14.4 	imes \frac{26}{M_{solute}}$. $M_{solute} = \frac{14.4 	imes 26}{2.304} = \frac{374.4}{2.304} = 162.5 \ g/mol$.

Bromoform has a normal freezing point of $7.734^{\circ} C$ and its $K_{f} = 14.4^{\circ} C / m$. $A$ solution of $2.60 \ g$ of an unknown substance in $100 \ g$ of bromoform freezes at $5.43^{\circ} C$. What is the molecular weight of the unknown substance?

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