Under identical conditions,which aqueous solu | vedclass

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

Question

(B) The freezing point depression $\Delta T_f$ is a colligative property,which depends on the molality $(m)$ of the solution. $\Delta T_f = K_f 	imes

Vedclass · Accepted Answer

$6 \ g$ urea in $100 \ g \ H_2O$ and $18 \ g$ glucose in $100 \ g \ H_2O$

Vedclass · Answer

(B) The freezing point depression $\Delta T_f$ is a colligative property,which depends on the molality $(m)$ of the solution. $\Delta T_f = K_f 	imes m$. For the freezing points to be equal,the molality of the two solutions must be equal. Molality $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}}$. For option $B$: Moles of urea $= \frac{6 \ g}{60 \ g/mol} = 0.1 \ mol$. Moles of glucose $= \frac{18 \ g}{180 \ g/mol} = 0.1 \ mol$. Since the mass of the solvent $(100 \ g \ H_2O)$ is the same for both,and the number of moles of solute is the same $(0.1 \ mol)$,the molality of both solutions is identical. Therefore,they will have the same freezing point.

Under identical conditions,which aqueous solutions have the same freezing point? (Molecular mass of urea $= 60 \ u$ and glucose $= 180 \ u$)

Similar Questions

Which of the following aqueous solutions has the highest freezing point?

At the freezing point of a solution containing a nonvolatile solute,which of the following are in equilibrium?

Calculate $\Delta T_{f}$ of aqueous $0.01 \ m$ formic acid if the van't Hoff factor is $1.1$. $[K_{f} = 1.86 \ K \ kg \ mol^{-1}]$ (in $K$)

Which of the following solutions exhibits the highest freezing point depression?

Calculate the molality of a solution of a non-volatile solute having a depression in freezing point of $0.93 \ K$ and a cryoscopic constant of the solvent of $1.86 \ K \ kg \ mol^{-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module