Question

How much pure alcohol must be added to 400 ml of a 15% solutions to make its strength 32%

Answer 1

Here is ur answer, dear...
_____________________⤵

➧ Volume of alcohol in 400 ml solution
➾ 0.15 × 400 ml

➧ Volume of alcohol in 400 ml solution
➾ 60 ml

➧ Volume of others liquid
➾ 340 ml

▶Let x ml of alcohol is added to make it 32% strong

➧ Then, volume of alcohol
➾ 60+x ml

➧ Volume of other liquid
➾ 340 ml

➧ Volume of solution
➾ 400 + x ml

➾ 0.32 × (400 + x) = 60 + x

➾ 128 + 0.32x = 60 + x

➾ 0.68x = 68

➾ 68x = 6800

➾ x = 100 ...✔

_________
Thanks...✊

Answer 2

ĀNSWĒR ⏬⏬⏬


$let \: the \: number \: of \: ml \: of \: alchol \: \\ = x \\ \\ given \: 15\%solution \: of \: 400ml \\ should \: be \: added \: with \: 100\% \: of \: \\ pure \: alchol. \\ \\ so.400(15\% ) + x(100\%) = \\ (400 + x)(32\%) \\ = 400 \times \frac{15}{100} + x = (400 + x) \times \frac{32}{100} \\ = 60 + x = 128 + ( \frac{8x}{25} ) \\ = x - \frac{8x}{25} = 128 - 60 \\ = \frac{17x}{25} = 68 \\ = x = \frac{(68 \times 25)}{17} \\ x = \frac{1700}{17} \\ \\ x = 100ml$

X:100ml


Thanks ✌✌✌✌