Question

How much pure alcohol must be added to 40 ml of a 15 percent solution to make its strength 32percents​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

${ \red{ \text{ \underline{ \purple{ \underline{Given \: data}}}}}:-}$

☆ Pure alcohol added to 40 ml of a 15% solution to make its strength 32%.

${ \red{ \text{ \underline{ \purple{ \underline{Solution}}}}}:-}$

Let, Quantity of alcohol in 40 ml solution be Q .

${ \sf{ \dashrightarrow{ \frac{Q}{40} \times 100 = 15}}}$

${ \sf{ \dashrightarrow{ Q = \frac{15 \times 40}{100} = 6}}}$

Pure alcohol added to 40 ml of a 15% solution be x. {to make its strength 32%.} so, here

${ \sf{ \dashrightarrow{ \frac{ 6 \: + \: x}{40 \: + \: x} \times 100 = 32}}}$

${ \sf{ \dashrightarrow{ \frac{ 600\: + \: 100x}{40 \: + \:100 x} = 32}}}$

${ \sf{ \dashrightarrow{ { 600\: + \: 100x} = 32 \: ({40 \: + \: x}) }}}$

${ \sf{ \dashrightarrow{ { 600\: + \: 100x} = {1280 \: + \:32 x} }}}$

${ \sf{ \dashrightarrow{ 68x} = {680} }}$

${ \sf{ \dashrightarrow{ x = \frac{680}{68} } }}$

${ \sf{ \dashrightarrow{ x = 10} \: ml }}$

Hence, 10 ml alcohol need to added in 15% alcohol to make solution of 32%.