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How many millilitres of 0.5 M H_(2)SO_(4) are needed to dissolve 0.5 g of copper (II) carbonate ? (At. Mass : H = 1, C = 12, O = 16, S = 32, Cu = 63.5) |
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Answer» Solution :`CuCO_(3)+H_(2)SO_(4)rarr CuSO_(4)+CO_(2)+H_(2)O` `"1 MOLE of "CuCO_(3)=63.5+12+48=123.5" G required "H_(2)SO_(4)="1 mole"` `THEREFORE 0.5 g CuCO_(3)" will required "H_(2)SO_(4)=(1)/(123.5)xx0.5"mole"=(1)/(247)"mole"` `"0.5 mole of 0.5 M "H_(2)SO_(4)" are present in 1000 mL"` `therefore""(1)/(247)" mole of 0.5 M "H_(2)SO_(4)" will be present in "(1000)/(0.5)xx(1)/(247)mL=8.1mL` |
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