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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%.
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Solution
Given: Vapour pressure of pure solvent octane `(p_1^0)` = p
Vapour pressure of the solution (p1) = 80% of p = 0.80 p
Molar mass of solute (M2) = 40 g mol−1
Mass of solvent octane (w1) = 114 g
Molar mass of solvent octane (M1) = 114 g mol−1
According to Raoult’s Law for a non-volatile solute:
`(p_1^0 - p)/p_1^0 = n_2/(n_1 + n_2)`
Number of moles of solvent octane (n1) = `114/114`
= 1 mol
Number of moles of solute (n2) = `w_2/40`
Substitute the values into Raoult’s Law formula:
`(p - 0.80 p)/p = (w_2/40)/(1 + w_2/40)`
`(0.20 p)/p = (w_2/40)/((40 + w_2)/40)`
0.2 = `w_2/(40 + w_2)`
0.2 × (40 + w2) = w2
8 + 0.2 w2 = w2
8 = w2 − 0.2 w2
8 = 0.8 w2
w2 = `8/0.8`
= 10 g
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