Question

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

Answer 1

Solution :

Given concentration of pure alcohol is 15% .

Amount of alcohol in the solution :

=> 15% of 400

=> 15/100 × 400

=> 15 × 4

=> 60 ml

Amount of solute :

=> 400 ml - 60 ml

=> 340 ml

Let the amount of alcohol added be x ml.

Required percentage of alcohol now :

=> ( 60 + x)/( 340 + x ) = 32/100

=> 100( 60 + x ) = 32( 340 + x)

=> 6000 + 100x = 10880 + 32x

=> 68x = 4880

=> x = 72 ml approx .

Thus, 72 ml of pure alcohol needs to be added to meet the given conditions.

This is the required answer .

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