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 o

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1. 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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