How much pure alcohol has to be added to 400 | vedclass

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(C) Initial volume of solution $= 400 	ext{ ml}$.Quantity of alcohol $= 15 \% 	ext{ of } 400 	ext{ ml} = \frac{15}{100} 	imes 400 = 60 	ext{ ml}$

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(C) Initial volume of solution $= 400 	ext{ ml}$.Quantity of alcohol $= 15 \% 	ext{ of } 400 	ext{ ml} = \frac{15}{100} 	imes 400 = 60 	ext{ ml}$.Quantity of water (which remains constant) $= 400 - 60 = 340 	ext{ ml}$.Let the amount of pure alcohol to be added be $x 	ext{ ml}$.New total volume $= 400 + x 	ext{ ml}$.New quantity of alcohol $= 60 + x 	ext{ ml}$.According to the problem,the new concentration is $32 \%$,so the water concentration is $100 \% - 32 \% = 68 \%$.Therefore,$68 \% 	ext{ of } (400 + x) = 340$.$0.68 	imes (400 + x) = 340$.$400 + x = \frac{340}{0.68} = 500$.$x = 500 - 400 = 100 	ext{ ml}$.

How much pure alcohol has to be added to $400 \text{ ml}$ of a solution containing $15 \%$ alcohol to change the concentration of alcohol in the mixture to $32 \%$?

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