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-

Question

Toppr Admin · Accepted Answer

Amount of acid in the 1st solution -0-25- of 200-50 g-Amount of acid in the 2nd solution -0-4- of-xA0- 300-120 g-Total amount of acid -50-120-170 g-xA0-Required concentration -170-200-300-xD7-100-34-

$200$ 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 $= \frac{170}{(200 + 300)} \times 100 = 34 %$

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 $= \frac{170}{(200 + 300)} \times 100 = 34 %$