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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$271$

Vedclass · Answer

(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 $= 273.15 \ K$]

Similar Questions

$0.440 \ g$ of a substance dissolved in $22.2 \ g$ of benzene lowered the freezing point of benzene by $0.567 \ ^oC$. Calculate the molecular mass of the substance. (Given: $K_f = 5.12 \ ^oC \ kg \ mol^{-1}$)

$A$ solution containing $62 \ g$ of ethylene glycol in $250 \ g$ of water is cooled to $-10 \ ^\circ C$. If $K_f$ for water is $1.86 \ K \ kg \ mol^{-1}$,the amount of water (in $g$) separated as ice is:

Mass of ethylene glycol (antifreeze) to be added to $18.6 \ kg$ of water to protect the freezing point at $-24^{\circ} C$ is . . . . . . $kg$ (Molar mass in $g \ mol^{-1}$ for ethylene glycol $= 62$,$K_{f}$ of water $= 1.86 \ K \ kg \ mol^{-1}$)

$2 \ g$ of a non-electrolyte solute (molar mass is $500 \ g \ mol^{-1}$) was dissolved in $57.3 \ g$ of xylene. If the freezing point depression constant $K_f$ of xylene is $4.3 \ K \ kg \ mol^{-1}$,then the depression in freezing point of xylene is.......... (in $K$)

Consider the given plots of vapour pressure $(VP)$ vs temperature $(T / K)$. Which amongst the following options is the correct graphical representation showing $\Delta T_f$,the depression in the freezing point of the solvent in a solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$40 \ g$ 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 $= 273.15 \ K$]