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Doubt

What is the volume of 50% (w/v) H2SO4 required for the liberation of 5.6l of hydrogen gas at STP on its reaction with magnesium.

Answer Explanation

W/v represents weight of solute in 100 ml of solution x 100.

Mg + H2SO4 --> MgSO4 + H2

5.6l of hydrogen = 5.6/22.4 = 0.25 moles of hydrogen. This is produced by 0.25 moles of sulphuric acid.

So mass of H2SO4 = 0.25 x 98 = 24.5 grams.

But w/v = 0.5 = w/100

So, weight in 100 ml = 50 grams.

Therefore, 50 grams are present in 100 ml, 24.5 grams are present in (100/50) x 24.5 = 49 ml of solution. So, the volume is 49 ml.
 

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