If the freezing point of a 5% (by mass) aqueo | vedclass

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

Question

(C) The depression in freezing point is given by $\Delta T_f = K_f 	imes m$,where $m$ is the molality.For a $5\%$ solution by mass,the mass of solute

Vedclass · Accepted Answer

$269.07$

Vedclass · Answer

(C) The depression in freezing point is given by $\Delta T_f = K_f 	imes m$,where $m$ is the molality.For a $5\%$ solution by mass,the mass of solute is $5 \ g$ and the mass of solvent is $95 \ g$.Molality $m = \frac{w_2 	imes 1000}{M_2 	imes w_1}$.Since $w_2$ and $w_1$ are the same for both cane sugar $(M_2 = 342 \ g/mol)$ and glucose $(M_2 = 180 \ g/mol)$,$\Delta T_f$ is inversely proportional to the molar mass $M_2$.$\Delta T_{f, 	ext{sugar}} = 273.15 - 271 = 2.15 \ K$.$\frac{\Delta T_{f, 	ext{glucose}}}{\Delta T_{f, 	ext{sugar}}} = \frac{M_{	ext{sugar}}}{M_{	ext{glucose}}} = \frac{342}{180} = 1.9$.$\Delta T_{f, 	ext{glucose}} = 1.9 	imes 2.15 = 4.085 \ K$.Freezing point of glucose solution $= 273.15 - 4.085 = 269.065 \ K \approx 269.07 \ K$.

If the freezing point of a $5\%$ (by mass) aqueous solution of cane sugar is $271 \ K$ and the freezing point of pure water is $273.15 \ K$,then the freezing point of a $5\%$ (by mass) aqueous solution of glucose will be .......... $K$.

Similar Questions

The freezing point of a $0.05 \ molal$ solution of a non-electrolyte in water is $(K_f = 1.86 \ K \ kg \ mol^{-1})$:

The freezing point of a $5\%$ (by mass) aqueous solution of sucrose is $271 \, K$ and the freezing point of pure water is $273.15 \, K$. The freezing point of a $5\%$ (by mass) aqueous solution of glucose is .......... $K$.

$CaCl_2$ is preferred over $NaCl$ for clearing ice on roads,particularly in very cold countries. This is because:

What is the cryoscopic constant of water if $5 \ g$ of glucose in $100 \ g$ of water causes a depression in freezing point of $2.15 \ K$? (Molar mass of glucose $= 180 \ g \ mol^{-1}$)

Bromoform has a normal freezing point of $7.734^{\circ} C$ and its $K_{f} = 14.4^{\circ} C / m$. $A$ solution of $2.60 \ g$ of an unknown substance in $100 \ g$ of bromoform freezes at $5.43^{\circ} C$. What is the molecular weight of the unknown substance?

Vedclass Test Series

Exam Paper Generator

Online Exam Module