• # question_answer The freezing point depression constant for water is $-1.86{{\,}^{o}}C\,{{m}^{-1}}.$If $5.00\,g\,N{{a}_{2}}S{{O}_{4}}$is dissolved in $45.0\,g\,{{H}_{2}}O,$the freezing point is changed by $-3.82{{\,}^{o}}C,$ Calculate the van't Hoff factor for $N{{a}_{2}}S{{O}_{4}}.$ A)  0.381         B)  2.63C)  2.05          D)  3.11

$\Delta T=\frac{100\times {{K}_{f}}\times w}{m.W}\times i$ $\Delta \Tau =0-(-3.82)=3.82{{\,}^{o}}C$ $3.82=\frac{1000\times 1.86\times 5\times i}{142\times 45}$ $i=2.63$