How many grams of water are required to add to 200 g of 98% sulfuric acid solution for obtaining a solution with a mass fraction of acid 20%
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Answer:
Explanation:
20% = Mass fraction of 1/5 0r 0.5
98% of 200 g sulfuric acid = 196 g
Number and moles of sulphuric is 196/98 = 2 moles
0.5 = 2/ 2+x
1+0.5x = 2
0.5x = 1
X= 2
= 2 moles of water
2×18
= 36 grams of water
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