40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is _____ K. (Nearest integer)
Question:
40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is _____ K. (Nearest integer)
$\left[\right.$ Given $: \mathrm{K}_{\mathrm{r}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} ;$ Density of water $=$
$1.00 \mathrm{~g} \mathrm{~cm}^{-3}$; Freezing point of water $\left.=273.15 \mathrm{~K}\right]$
Solution:
molality $=\frac{\left(\frac{40}{180}\right) \mathrm{mol}}{0.2 \mathrm{Kg}}=\left(\frac{10}{9}\right)$ molal
$\Rightarrow \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{f}}^{\prime}=1.86 \times \frac{10}{9}$
$\Rightarrow \mathrm{T}_{\mathrm{f}}^{\prime}=273.15-1.86 \times \frac{10}{9}$
$=271.08 \mathrm{~K}$
$\simeq 271 \mathrm{~K}$ (nearest-integer)