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# What volume of 98% sulphuric acid (in ml) should be mixed with water to obtain 200 mL of 15% solution of sulphuric acid by weight? Given density of H2O=1.00 g cm−3, sulphuric acid (98%) =1.88 g cm−3 and sulphuric acid (15%) =1.12 g cm−3.

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## Correct Answer - V1=18.2mL 1. In the case of dilution, the simplest way is to determine the normality of the 98% and 15% H2SO4 . 2. For 98% H2SO4, the volume of H2SO4 in 100g solution is = 1001.88cm3. 3. Let normality of 98% H2SO4 is N1, and N1=1000×WE×V , where W is the weight of H2SO4 (which is 98g), E is the equivalent weight of H2SO4 (which is 49g). V is the volume that we have calculated above that is 1001.88cm3 3. N1=1000×9849×1001.88 = 37.6N.4. For 15% H2SO4, the volume of H2SO4 in 100g solution is =1001.12cm3 5. Let normality of 15% H2SO4 is N2 and N2=1000×wE1×v where w is the weight of H2SO4(which is 98g) , E is the equivalent weight of H2SO4 which is 49g.6.N2=1000×9849×1001.12 = 3.43N 6. As we know that N1V1=N2V2, 7. (37.6)×V1=(3.43)×200 8. V1=18.2mL  Suggest Corrections  0      Similar questions  Related Videos   Volume of Gases and Number of Moles
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