The quantity of water needed to reduce 9 ml | vedclass

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(C) Initial volume of shaving lotion = $9 \ ml$.Alcohol concentration = $50 \%$.Amount of alcohol = $9 	imes 0.50 = 4.5 \ ml$.Let the total volume of

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) Initial volume of shaving lotion = $9 \ ml$.Alcohol concentration = $50 \%$.Amount of alcohol = $9 	imes 0.50 = 4.5 \ ml$.Let the total volume of the new lotion be $V \ ml$ after adding water.The amount of alcohol remains constant at $4.5 \ ml$.In the new lotion,the concentration of alcohol is $30 \%$.So,$30 \%$ of $V = 4.5 \ ml$.$0.30 	imes V = 4.5$.$V = \frac{4.5}{0.30} = 15 \ ml$.The quantity of water to be added = New total volume - Initial volume.Water added = $15 \ ml - 9 \ ml = 6 \ ml$.

The quantity of water needed to reduce $9 \ ml$ of shaving lotion containing $50 \%$ alcohol to a lotion containing $30 \%$ alcohol is: (in $ml$)

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