Question:
Calculate the mass of a non-volatile solute (molar mass \(40 g mol^{–1}\)) which should be dissolved in 114 g octane to reduce its vapour pressure to \(80\%\)
Calculate the mass of a non-volatile solute (molar mass \(40 g mol^{–1}\)) which should be dissolved in 114 g octane to reduce its vapour pressure to \(80\%\)
Updated On: Sep 28, 2023
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Solution and Explanation
The correct answer is: \(8 g\)
Let the vapour pressure of pure octane be \(p^o_1.\)
Then, the vapour pressure of the octane after dissolving the non-volatile solute is \(\frac{80}{100} p^o_1 = 0.8p^o_1.\)
Molar mass of solute, \(M_2 = 40 g mol^{- 1}\)
Mass of octane, \(w_1 = 114 g\)
Molar mass of octane, \((C_8H_{18}), M_1 = 8 × 12 + 18 × 1 = 114 g mol^{- 1}\)
Applying the relation,
\(\frac{ (p^o_1-p_1)}{ p^o_1}= \frac{(w_2\times M_1)}{(M_2 \times w_1)}\)
\(⇒ \frac{(p^o_1-0.8p_1)}{ p^o_1}= \frac{(w_2\times 114)}{(40 \times 114)}\)
\(⇒\frac{0.2p^o_1}{p^o_1} = \frac{w_2}{40}\)
\(⇒0.2= \frac{w_2}{40}\)
\(⇒w_2 = 8g\)
Hence, the required mass of the solute is \(8 g\).
Let the vapour pressure of pure octane be \(p^o_1.\)
Then, the vapour pressure of the octane after dissolving the non-volatile solute is \(\frac{80}{100} p^o_1 = 0.8p^o_1.\)
Molar mass of solute, \(M_2 = 40 g mol^{- 1}\)
Mass of octane, \(w_1 = 114 g\)
Molar mass of octane, \((C_8H_{18}), M_1 = 8 × 12 + 18 × 1 = 114 g mol^{- 1}\)
Applying the relation,
\(\frac{ (p^o_1-p_1)}{ p^o_1}= \frac{(w_2\times M_1)}{(M_2 \times w_1)}\)
\(⇒ \frac{(p^o_1-0.8p_1)}{ p^o_1}= \frac{(w_2\times 114)}{(40 \times 114)}\)
\(⇒\frac{0.2p^o_1}{p^o_1} = \frac{w_2}{40}\)
\(⇒0.2= \frac{w_2}{40}\)
\(⇒w_2 = 8g\)
Hence, the required mass of the solute is \(8 g\).
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