At room temperature,a dilute solution of urea | vedclass

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

Question

(D) The lowering of vapour pressure is given by the formula: $\Delta p = p^o \cdot X_{solute}$Given:$p^o = 35 \ mm \ Hg$$w_{urea} = 0.60 \ g$,$M_{urea

Vedclass · Accepted Answer

$0.017$

Vedclass · Answer

(D) The lowering of vapour pressure is given by the formula: $\Delta p = p^o \cdot X_{solute}$Given:$p^o = 35 \ mm \ Hg$$w_{urea} = 0.60 \ g$,$M_{urea} = 60 \ g \ mol^{-1}$$w_{water} = 360 \ g$,$M_{water} = 18 \ g \ mol^{-1}$Calculate moles of solute $(n)$ and solvent $(N)$:$n = \frac{0.60}{60} = 0.01 \ mol$$N = \frac{360}{18} = 20 \ mol$Calculate mole fraction of solute $(X_{solute})$:$X_{solute} = \frac{n}{n + N} = \frac{0.01}{0.01 + 20} = \frac{0.01}{20.01} \approx 0.00049975$Calculate lowering of vapour pressure $(\Delta p)$:$\Delta p = 35 	imes 0.00049975 \approx 0.01749 \ mm \ Hg$Rounding to the nearest value,we get $0.017 \ mm \ Hg$.

At room temperature,a dilute solution of urea is prepared by dissolving $0.60 \ g$ of urea in $360 \ g$ of water. If the vapour pressure of pure water at this temperature is $35 \ mm \ Hg$,the lowering of vapour pressure will be: .............. $mm \ Hg$ (molar mass of urea $= 60 \ g \ mol^{-1}$)

Similar Questions

For two components $A$ and $B$,$P_A^0 : P_B^0 = 1:2$ and $X_A : X_B = 1:2$. The mole fraction of $A$ in the vapor phase is:

The vapour pressure of a dilute aqueous solution of glucose is $750 \ mm$ of mercury at $373 \ K$. The mole fraction of solvent is

Vapour pressure of solvent $A$ is $0.90 \text{ atm}$. When a non-volatile solute is added,the vapour pressure drops to $0.60 \text{ atm}$. What is the mole fraction of $A$ in the solution?

The vapour pressure of water is $12.3 \, kPa$ at $300 \, K$. Calculate the vapour pressure of a $1 \, molal$ solution of a non-volatile solute in it.

Vedclass Test Series

Exam Paper Generator

Online Exam Module