40 g of glucose (Molar mass = 180 g mol^{- | vedclass

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(A) Step $1$: Calculate the number of moles of glucose.$n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{40 \ g}{180 \ g \ mol^{-1}} = \frac{2}{9} \

Vedclass · Accepted Answer

$271$

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(A) Step $1$: Calculate the number of moles of glucose.$n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{40 \ g}{180 \ g \ mol^{-1}} = \frac{2}{9} \ mol \approx 0.222 \ mol$.Step $2$: Calculate the mass of the solvent (water).Density $= 1.00 \ g \ cm^{-3}$,so $200 \ mL$ of water $= 200 \ g = 0.2 \ kg$.Step $3$: Calculate the molality $(m)$ of the solution.$m = \frac{n_{	ext{solute}}}{W_{	ext{solvent (kg)}}} = \frac{2/9 \ mol}{0.2 \ kg} = \frac{10}{9} \ m \approx 1.11 \ m$.Step $4$: Calculate the depression in freezing point $(\Delta T_{f})$.$\Delta T_{f} = K_{f} 	imes m = 1.86 \ K \ kg \ mol^{-1} 	imes \frac{10}{9} \ mol \ kg^{-1} \approx 2.067 \ K$.Step $5$: Calculate the freezing point of the solution $(T_{f}^{\prime})$.$T_{f}^{\prime} = T_{f} - \Delta T_{f} = 273.15 \ K - 2.067 \ K = 271.083 \ K$.Rounding to the nearest integer,we get $271 \ K$.

$40 \ g$ of glucose (Molar mass $= 180 \ g \ mol^{-1}$) is mixed with $200 \ mL$ of water. The freezing point of the solution is $..... \ K$. (Nearest integer)
[Given : $K_{f} = 1.86 \ K \ kg \ mol^{-1};$ Density of water $= 1.00 \ g \ cm^{-3};$ Freezing point of water $= 273.15 \ K$]

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$40 \ g$ of glucose (Molar mass $= 180 \ g \ mol^{-1}$) is mixed with $200 \ mL$ of water. The freezing point of the solution is $..... \ K$. (Nearest integer)[Given : $K_{f} = 1.86 \ K \ kg \ mol^{-1};$ Density of water $= 1.00 \ g \ cm^{-3};$ Freezing point of water $= 273.15 \ K$]