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Q.

0.257 g of a metal when treated with excess dil.H2SO4 liberated 120 mL of H2 at NTP. Calculate the equivalentweight of the metal.

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answer is 23.99.

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Detailed Solution

wt. of metal =0.257g; VNTP=120mL We know that : Eq. wt. of a metal = wt. of metal VNTP of H2×11200=0.257g×11200mL120mL=23.99g Ans.

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