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Question
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
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Solution
Here, the vapour pressure of the solution at the normal boiling point (ps) = 1.004 bar
Vapour pressure of pure water at normal boiling point (p°) = 1 atm
= 1.013 bar
Let the mass of the solution = 100 g
Mass of solute (w2) = 2 g
Mass of solvent (water), (w1) = 100 − 2
= 98 g
Molar mass of solvent (water), (M1) = 18 g mol−1
According to Raoult’s law,
`(p^circ - p_s)/p^circ = n_2/(n_1 + n_2) = n_2/n_1`
= `(w_2//M_2)/(w_1//M_1) = w_2/M_2 xx M_1/w_1`
⇒ `(1.013 - 1.004)/1.013 = (2 xx 18)/(M_2 xx 98)`
⇒ `0.009/1.013 = (2 xx 18)/(M_2 xx 98)`
⇒ M2 = `(1.013 xx 2 xx 18)/(0.009 xx 98)`
= 41.35 g mol−1
Hence, the molar mass of the solute is 41.35 g mol−1.
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