Question

(c)
212,
25. How much pure alcohol should be added to 600
ml. of a 15% solution to make its strength 32%​

Answer 1

Answer:

150 ml alcohol must be added to make 32 % strength of the solution.

Step-by-step explanation:

Let x ml of pure alcohol be added to 600 ml of a 15% solution to make it strength 32%. Here, 15% solution means that there is 15 ml pure alcohol in a solution of 100 ml.

Now, quantity of alcohol in 100 ml solution=15ml

Therefore,quantity of alcohol in 1 ml solution = 15/100ml

quantity of alcohol in 600 ml solution = 15/100×600ml

=90ml

Total quantity of the solution =(600+x) ml

total quantity of alcohol in (600+x) ml solution= (90+x) ml

therefore, quantity of alcohol in 1 ml = 90+x/600+x ml

quantity of alcohol in 100 ml = 90+x/600+x×100ml

So, strength of the solution= (90+x/600+x)×100 percent

But, the strength of the solution is given as 32%.

therefore, 90+x/600+x×100 = 32

=) 100(90+x) = 32(600+x)

=) 9000+ 100x = 19200+32x

=) 100x- 32x = 19200-9000

=) 68x = 10200

=) x = 10200/68

=) x = 150.

Mark me as brainliest.