$40g$ of glucose (Molar mass $=180)$ is mixed with $200mL$ of water. The freezing point of solution is ________$K$. (Nearest integer)
$\text{[Given}:K_f=1.86Kmo{l}^{-1}\text{Density of water}=1.00gc{m}^{-3};\text{Freezing point water}=273.15K]$

Freezing point depression of a solution is given by,
$△T_f=i\times K_f\times m$
where,
$△T_f$ is the depression in freezing point.
$K_f$ is molal depression constant.
m is molality of solution.
$\text{Van't hoff factor, i}=1\text{for glucose})$
$\text{Molality (m)}=\frac{\frac{w_{\text{solute}}}{M}\times 1000}{W_{\text{solvent}}}$
where,
$w$ is weight of solute
$W$ is weight of solvent
$M$ is molar mass of the solute

$\Delta T_f=1\times 1.86\times \frac{40\times 1000}{180\times 200}(\frac{\text{Mass of water}=200g}{asd=1g/mL})$

$=2.06$


$\therefore$ Freezing point $=T_o-\Delta T_f$
$T_o$ is freezing point of water
$=273.15-2.06$

$=271.09K$

$=271K$