Chemistry - Liquid Solution Question with Solution | TestHub

$180\times 0.5\mathrm{g}$ glucose present per 1000 g of ${\mathrm{H}}_2\mathrm{O}(\ell )$

90 g glucose present per 1090 g of glucose solution

545 g solution contain $\frac{90}{1090}\times 545=45$ g glucose

$\therefore$ Initial weight of water $=545-45=500\mathrm{g}$

$\Delta {\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\cdot {\mathrm{m}}_{\text{final }};3.6=1.8\times {\mathrm{m}}_{\text{fina }}$

${\mathrm{m}}_{\text{final }}=2$

$2\times 180\mathrm{g}$ glucose present with $1000\mathrm{g}{\mathrm{H}}_2\mathrm{O}$

45 g glucose present with $\frac{1000}{2\times 180}\times 45=125\mathrm{g}{\mathrm{H}}_2\mathrm{O}$

$\therefore$ Final wt. of water $=125\mathrm{g}$

Wt. of ice formed $=500-125=375\mathrm{g}$

$\%$ Wt. of water converted into ice $=\frac{375}{500}\times 100=75$