Question

Calculate the freezing point of solution when 1.9 g of $\text{MgC}{\text{l}}_{\text{2}}$ $\text{(M - 95g}\text{mo}{\text{l}}^{\text{- 1}})$ was dissolved in 50g of water, assuming $\text{MgC}{\text{l}}_{\text{2}}$ undergoes complete ionisation. $\left[{\text{K}}_f\text{for water= 1}\text{.86}\text{K}\text{Kg}\text{mo}{\text{l}}^{-1}\right]$

Answer 1

$\Delta T_f=i{K}_{f}m$

$\Delta T_f=\frac{\left(3\right)\left(1.86\right)\times 2}{5}$

$\Delta T_f=2.232$

$T_{initial}-T_{final}=2.232$

$T_{final}=T_{initial}-2.232$

$T_{final}=0^{o}C-{2.232}^{o}C$

$T_{final}=-{2.232}^{o}C$

Note:

$K_f=1.86KKg.mo{l}^{-1}$

$i=3[MgC{l}_2\rightleftharpoons M{g}^{2+}+2C{l}^{-}i=\frac{2+1}{1}=3]$

$m=\frac{Givenmass}{Molecularmass}\times \frac{1}{weightofsolvent\left(kg\right)}$

$m=\frac{1.9}{95}\times \frac{1}{50}\times 1000mol/kg$

$m=\frac{20}{50}=\frac{2}{5}mol/kg$