how much pure alcohol must be added to 400 ml of a 15% solution to make it is strength 32%
Answers
400ml has 15% alcohol i.e 60ml . This means we have 340ml of water or the liquid with which alcohol is mixed
Let V be the volume of alcohol added to make the strength 32%
then V/340+V = 32/100. solving this we get V as 160ml
so we need to add 160 -60 = 100ml of alcohol
so the alcohol already present in solution is 400*15/100= 60 ml
let the volume of alcohol added be x ml so the final volume of solution becomes 400+x
now (400+x)*32/100 = 60 + x
which simplifies to x=100 ml
Hi mate,
Solution :
=> 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 ml ...
Answer : 100 ml pure alcohol must be added to 400ml of a 15% solution to make it's strength 32%
i hope it helps you