Calculate the freezing point of a solution pr | vedclass

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(B) $1$. Molar mass of glucose $(C_6H_{12}O_6) = (6 	imes 12) + (12 	imes 1) + (6 	imes 16) = 180 \ g \ mol^{-1}$.
$2$. Number of moles of glucose

Vedclass · Accepted Answer

$273.11$

Vedclass · Answer

(B) $1$. Molar mass of glucose $(C_6H_{12}O_6) = (6 	imes 12) + (12 	imes 1) + (6 	imes 16) = 180 \ g \ mol^{-1}$.
$2$. Number of moles of glucose = $\frac{1.8 \ g}{180 \ g \ mol^{-1}} = 0.01 \ mol$.
$3$. Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.01 \ mol}{0.5 \ kg} = 0.02 \ m$.
$4$. Depression in freezing point $(\Delta T_f) = K_f 	imes m = 1.86 \ K \ kg \ mol^{-1} 	imes 0.02 \ mol \ kg^{-1} = 0.0372 \ K$.
$5$. Freezing point of solution $(T_f) = T_f^{\circ} - \Delta T_f = 273.15 \ K - 0.0372 \ K = 273.1128 \ K \approx 273.11 \ K$.

Column-$I$ (Various solutions)	Column-$II$ (Freezing point)
$a$. $0.1 \, M \ BaCl_2$ solution	$p$. $271 \, K$
$b$. $0.1 \, M \ NaCl$ solution	$q$. $270 \, K$
$c$. $0.1 \, M \ K_3[Fe(CN)_6]$ solution	$r$. $268 \, K$
$d$. $0.1 \, M \ Al_2(SO_4)_3$ solution	$s$. $269 \, K$

Calculate the freezing point of a solution prepared by dissolving $1.8 \ g$ of glucose $(C_6H_{12}O_6)$ in $500 \ g$ of water. The $K_f$ value for water is $1.86 \ K \ kg \ mol^{-1}$. (in $K$)

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