A solution is prepared by adding 1 mole ethyl alcohol in 9 mole water. The mass percent of solute in the solution is

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Solution

Q.
A solution is prepared by adding 1 mole ethyl alcohol in 9 mole water. The mass percent of solute in the solution is ______ (Integer Answer)

(Given: Molar mass in g ${mol}^{- 1}$ Ethyl alcohol : 46, water : 18) [2024]

Ans.
(22)

$W_{C_{2} H_{5} O H} = n_{C_{2} H_{5} O H} \times M_{C_{2} H_{5} O H} = 1 mol \times 46 {g mol}^{- 1} = 46 g$

$W_{H_{2} O} = n_{H_{2} O} \times M_{H_{2} O} = 9 mol \times 18 {g mol}^{- 1} = 162 g$

$Mass percent of solute = \frac{W_{solute}}{W_{solution}} \times 100$

$= \frac{W_{C_{2} H_{5} O H}}{W_{H_{2} O} + W_{C_{2} H_{5} O H}} \times 100$

$= \frac{46}{162 + 46} \times 100 = 22.11 % \approx 22 %$

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