In an experiment,$1 \ g$ of a non-volatile solute was dissolved in $100 \ g$ of acetone (molar mass $= 58 \ g/mol$) at $298 \ K$. The vapour pressure of the solution was found to be $192.5 \ mm \ Hg$. The molecular weight of the solute is (vapour pressure of pure acetone $= 195 \ mm \ Hg$). (in $.24$)
- A$25$
- B$35$
- C$45$
- D$55$
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Similar Questions
Write the formula which expresses the relationship between the total pressure of a solution $(P_{total})$ and the mole fraction $(x_2)$ of the components,according to Raoult's law.
Easy
View Solution$X$ is a non-volatile solute and $Y$ is a volatile solvent. The following vapour pressures are observed by dissolving $X$ in $Y$ at different concentrations:
| $X / \text{mol L}^{-1}$ | $Y / \text{mm of Hg}$ |
| :--- | :--- |
| $0.10$ | $p_1$ |
| $0.25$ | $p_2$ |
| $0.01$ | $p_3$ |
The correct order of vapour pressures is:
| $X / \text{mol L}^{-1}$ | $Y / \text{mm of Hg}$ |
| :--- | :--- |
| $0.10$ | $p_1$ |
| $0.25$ | $p_2$ |
| $0.01$ | $p_3$ |
The correct order of vapour pressures is:
What is the molar mass of a solute when $2.3 \ g$ of a non-volatile solute is dissolved in $46 \ g$ of benzene at $30^{\circ} C$? (Relative lowering of vapour pressure is $0.06$ and the molar mass of benzene is $78 \ g \ mol^{-1}$)
An aqueous solution of methanol in water has a vapour pressure:
Easy
View SolutionWhat 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?
Medium
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