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

$A$ solution contains a $1:4$ molar ratio of pentane and hexane. At $20^{\circ}C$,the vapor pressure of pure pentane is $440 \ mm \ Hg$ and that of hexane is $120 \ mm \ Hg$. What is the mole fraction of pentane in the vapor phase?

Which of the following statements is incorrect?

Calculate the vapour pressure of a solution containing a mixture of $2 \ mol$ of volatile liquid $A$ and $3 \ mol$ of volatile liquid $B$ at room temperature. $(P_{A}^{\circ} = 420 \ mm \ Hg, P_{B}^{\circ} = 610 \ mm \ Hg)$ (in $mm \ Hg$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module