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Calculate the molality and molarity of a solution made by mixing equal volumes of 30% by weight of H_(2)SO_(4) (density =1.218g//mL) and 70% by weight of H_(2)SO_(4) (density =1.610g//mL) |
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Answer» Solution :Suppose that the solution contains 100mL of each variety of `H_(2)SO_(4)`. Total volume is THEREFORE `200ML` or `0.02` litre WT.of `100mL` of `H_(2)SO_(4)` solution `(30%)=1.218xx100` `=121.8g` and wt.of `100mL` of `H_(2)SO_(4)` solution `(70%)=1.610xx100` `=161g` Wt. of `H_(2)SO_(4)(30%)=121.8xx(30)/(100)=36.54g` Wt. of `H_(2)SO_(4)(70%)=161xx(70)/(100)=112.7g` Total wt of `H_(2)SO_(4)` (solute) =`36.54+112.7=149.24g` `:.` wt of `H_(2)O` (solvent) = wt.of solution `-` wt.of solute `=(121.8+161)-149.24` `=133.56g` Moles of `H_(2)SO_(4)=(149.24)/(8)=1.5228` (mol wt of `H_(2)SO_(4)=98`) MOLALITY `=(1.5228)//(133.56)xx1000=11.4m` Molarity `=(1.5228)/(0.2)=7.6M` |
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