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)
(Given : Molar mass in g mol$^{-1}$ Ethyl alcohol : 46, water : 18)
Correct Answer: 22
Approach Solution - 1
We are given the mass of ethyl alcohol and water, and we need to calculate the percentage by mass of ethyl alcohol in the solution:
Mass of ethyl alcohol = 1 mole × MM
$$ \Rightarrow 46 \, \text{g} $$Mass of water = 9 moles × MM
$$ \Rightarrow 162 \, \text{g} $$The percentage by mass of ethyl alcohol is given by:
$$ \% \, \text{by mass of ethyl alcohol} = \frac{46}{162 + 46} \times 100 $$ $$ \Rightarrow 22\% \, (\text{approx.}) $$Thus, the percentage by mass of ethyl alcohol is approximately 22%.
Approach Solution -2
Step 1: Calculate the mass of ethyl alcohol and water
\[ \text{Mass of ethyl alcohol} = 1 \, \text{mole} \times 46 \, \text{g mol}^{-1} = 46 \, \text{g}. \]
\[ \text{Mass of water} = 9 \, \text{mole} \times 18 \, \text{g mol}^{-1} = 162 \, \text{g}. \]
Step 2: Calculate total mass of solution
\[ \text{Total mass of solution} = \text{Mass of ethyl alcohol} + \text{Mass of water} = 46 + 162 = 208 \, \text{g}. \]
Step 3: Calculate mass percent of ethyl alcohol
\[ \text{Mass percent of ethyl alcohol} = \frac{\text{Mass of ethyl alcohol}}{\text{Total mass of solution}} \times 100. \]
\[ \text{Mass percent} = \frac{46}{208} \times 100 = 22.11\%. \]
Final Answer: 22
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