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