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  4. 9.8 g of H2SO4 is present in 2 L of a solution. The molarity of the solution is

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Q. 9.8 g of $ {{H}_{2}}S{{O}_{4}} $ is present in 2 L of a solution. The molarity of the solution is

CMC MedicalCMC Medical 2010

Solution:

Amount of $ {{H}_{2}}S{{O}_{4}}=9.8\,g $ Volume of solution = 2 L Moles of $ {{H}_{2}}S{{O}_{4}}=\frac{\text{mass}\,\text{of}\,{{H}_{2}}S{{O}_{4}}}{\text{molar}\,\text{mass}} $ $ =\frac{9.8\,g}{98\,g\,mo{{l}^{-1}}} $ ( $ \because $ Molar mass of $ {{H}_{2}}S{{O}_{4}} $ = 98) Molarity $ =\frac{\text{moles}\,\text{of}\,{{H}_{2}}S{{O}_{4}}}{\text{volume}\,\text{of}\,\text{solution}} $ $ =\frac{0.1\,\,mol}{2\,L} $ $ =0.05\,mol\,{{L}^{-1}} $ or $ 0.05\,M $