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How many molecules of water of crystallisationarepresentin1.648 gof copper sulphate (CuSO_(4).5H_(2)O) ? |
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Answer» Solution :Gram molecular mass of `CuSO_(4).5H_(2)O = 63.55 + 32.06 + (4 xx 16.0) + 5 xx (2 xx 1.008 + 16.0) = 249.69 g` This mass contains `6.022 xx 10^(23)`molecules of `CuSO_(4).5H_(2)O` `therefore` Total number of molecules of `CuSO_(4).5H_(2)O` present in 1.648 g `= (6.022 xx 10^(23))/(249.69)xx 1.648` `=3.975 xx 10^(21)` Since, each molecule of `CuSO_4.5H_2O` contains 5 molecules of water of crystallisation, therefore, total number of water molecules present in the given sample `3.975 xx 10^(21) xx 5 = 1.987 xx 10^(22)` Hence, 1.648 g of copper sulphate contain `1.987 xx 10^22` molecules of water of crystallisation. |
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