When 0.01 mol of sugar is dissolved in 100 | vedclass

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(D) The depression in freezing point is given by the formula $\Delta T_f = m 	imes K_f$,where $m$ is the molality of the solution and $K_f$ is the cr

Vedclass · Accepted Answer

$2.4$

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(D) The depression in freezing point is given by the formula $\Delta T_f = m 	imes K_f$,where $m$ is the molality of the solution and $K_f$ is the cryoscopic constant of the solvent.For the first case: $m_1 = \frac{0.01 \ mol}{100 \ g} 	imes 1000 \ g/kg = 0.1 \ mol/kg$.Given $\Delta T_{f1} = 0.40 \ ^oC$,we have $0.40 = 0.1 	imes K_f$,which gives $K_f = 4 \ ^oC \ kg/mol$.For the second case: $m_2 = \frac{0.03 \ mol}{50 \ g} 	imes 1000 \ g/kg = 0.6 \ mol/kg$.Now,calculating the new depression in freezing point: $\Delta T_{f2} = m_2 	imes K_f = 0.6 	imes 4 = 2.4 \ ^oC$.

When $0.01 \ mol$ of sugar is dissolved in $100 \ g$ of a solvent,the depression in freezing point is $0.40 \ ^oC$. When $0.03 \ mol$ of glucose is dissolved in $50 \ g$ of the same solvent,the depression in freezing point will be $......... \ ^oC$.

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