A solution containing 7.5 g of urea (molar m | 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 of the solution. Since both solutions have the

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(C) The depression in freezing point is given by $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution. Since both solutions have the same freezing point and the same amount of solvent ($1 \ kg$ of water),their molalities must be equal. Molality $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}}$. For urea: $	ext{moles} = \frac{7.5 \ g}{60 \ g \ mol^{-1}} = 0.125 \ mol$. For solute $X$: $	ext{moles} = \frac{15 \ g}{M_X}$,where $M_X$ is the molar mass of $X$. Equating the moles (since solvent mass is the same): $0.125 = \frac{15}{M_X}$. $M_X = \frac{15}{0.125} = 120 \ g \ mol^{-1}$.

$A$ solution containing $7.5 \ g$ of urea (molar mass $= 60 \ g \ mol^{-1}$) in $1 \ kg$ of water freezes at the same temperature as another solution containing $15 \ g$ of solute $X$,in the same amount of water. The molar mass of $X \ (g \ mol^{-1})$ is

Similar Questions

The mass of urea required to be dissolved in $500 \ mL$ of water to produce a depression in freezing point of $0.186 \ ^oC$ is ........ $g$. $(K_f = 1.86 \ ^oC/m)$

What happens to the freezing point of benzene when a small quantity of naphthalene is added to it?

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 $= 180 \ g \ mol^{-1}$)

$A$ solution of a nonvolatile solute is obtained by dissolving $15 \ g$ in $200 \ mL$ of water,which has a depression in freezing point of $0.75 \ K$. Calculate the molar mass of the solute if the cryoscopic constant of water is $1.86 \ K \ kg \ mol^{-1}$.

At $T$ $(K)$,$x \ g$ of a non-volatile solid (molar mass $78 \ g \ mol^{-1}$) when added to $0.5 \ kg$ water,lowered its freezing point by $1.0^{\circ} C$. What is $x$ (in $g$)? ($K_{f}$ of water at $T$ $(K)$ = $1.86 \ K \ kg \ mol^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module