$200$ g of $25 %$ sulphuric acid solution was added to $300$ g of $40 %$ sulphuric acid solution. Find the concentration of the acid in the mixture.

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Solution

Amount of acid in the 1st solution $= 0.25 %$ of $200 = 50$ g.
Amount of acid in the 2nd solution $= 0.4 %$ of $300 = 120$ g.
Total amount of acid $= 50 + 120 = 170$ g.
Required concentration $= \frac{170}{(200 + 300)} \times 100 = 34 %$

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200 g of 25% sulphuric acid solution was added to 300 g of 40% sulphuric acid solution. Find the concentration of the acid in the mixture.

Amount of acid in the 1st solution =0.25% of 200=50 g.Amount of acid in the 2nd solution =0.4% of 300=120 g.Total amount of acid =50+120=170 g. Required concentration =170(200+300)×100=34%

$200$ g of $25 %$ sulphuric acid solution was added to $300$ g of $40 %$ sulphuric acid solution. Find the concentration of the acid in the mixture.

Amount of acid in the 1st solution $= 0.25 %$ of $200 = 50$ g.
Amount of acid in the 2nd solution $= 0.4 %$ of $300 = 120$ g.
Total amount of acid $= 50 + 120 = 170$ g.
Required concentration $= \frac{170}{(200 + 300)} \times 100 = 34 %$