How many grams of ammonium sulphate will be required to prepare 6.8 gram of ammonia by heating ammonium sulphate with caustic soda solution? (N=14, O=16, S=32)
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Answer:
198.12 g
Explanation:
0.25 L of a solution with a molarity of 6M has 6*0.25 = 1.5 moles of the solute.
The molar mass of ammonium sulfate is 132.14 g/mole. The mass of 1.5 moles is 132.14*1.5 = 198.21 g.
Therefore 198.12 g of ammonium sulfate are required to make 0.25 L of a solution with a concentration of 6M.
Hope this helps you! ^-^
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