how much pure alcohol be added to 400 ml of a 15% solution to make it 10 32%
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Let the required number of milliliters of alcohols be x ml.Given 15 % of 400 ml should be added with 100 % of pure alcohol.Hence 400(15 %) + x(100 %) = (400+x) (32 %)⇒ 400 × 15/100 + x = (400 + x) × (32/100)⇒ 60 + x = 128 + (8x/25)⇒ x - 8x/25 = 128 - 60⇒ 17x/25 = 68⇒ x = (68 × 25)/17⇒ x = 1700/17x = 100 ml
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#SingodiyaBoy..
Hope It Help's..
#SingodiyaBoy..
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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
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