What is the mass in g of a non-volatile solut | vedclass

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

Question

(B) According to Raoult's Law for a non-volatile solute,the relative lowering of vapour pressure is given by: $\frac{P^0 - P_S}{P^0} = X_{	ext{solute

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) According to Raoult's Law for a non-volatile solute,the relative lowering of vapour pressure is given by: $\frac{P^0 - P_S}{P^0} = X_{	ext{solute}}$.Given that the vapour pressure is reduced to $80 \%$ of its original value,$P_S = 0.8 P^0$. Therefore,$\frac{P^0 - 0.8 P^0}{P^0} = \frac{0.2 P^0}{P^0} = 0.2$.Let $W$ be the mass of the solute. The number of moles of solute $n = \frac{W}{40}$.The molar mass of octane $(C_8H_{18})$ is $8 	imes 12 + 18 	imes 1 = 114 \ g/mol$. The moles of octane $N = \frac{57}{114} = 0.5 \ mol$.Using the formula $\frac{n}{n + N} = 0.2$,we get $\frac{W/40}{W/40 + 0.5} = 0.2$.$W/40 = 0.2(W/40 + 0.5) \implies W/40 = 0.005W + 0.1$.$0.025W - 0.005W = 0.1 \implies 0.02W = 0.1 \implies W = 5 \ g$.

What is the mass in $g$ of a non-volatile solute with a molecular weight of $40$ that should be dissolved in $57 \ g$ of octane to reduce its vapour pressure to $80 \%$ of its original value?

Similar Questions

The graph between vapour pressure and mole fraction of the solvent for an ideal solution is:

The relative lowering of vapour pressure produced by dissolving $71.5 \ g$ of a substance in $1000 \ g$ of water is $0.00713$. The molecular weight of the substance will be:

Calculate the mole fraction of solute in a solution if the vapour pressure of the solution and the pure solvent are $27 \text{ mm Hg}$ and $30 \text{ mm Hg}$ respectively at $298 \text{ K}$.

Which of the following can be measured by the Ostwald-Walker dynamic method?

Which solution will show the maximum vapour pressure at $300 \ K$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module