3.1 g of a metal was treated with excess of HCl. As a result, 1100.4 mL of H2was collected at 15°C and 750 mm pressure. Calculate the equivalent weight of the metal. Aqueous tension at 15°C is 12.5 mm; wt. of 1 mL of H2 NTP=0.00009 g.

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answer is 33.7.

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Detailed Solution

Wt.ofmetal =3.1gP1=750mm−12.5mm=737.5mm; V1=1100.4mL;T1=15+273=288K; At. NTP, P2=760mm,V2=?,T2=273K.Using gas equation, We have: P1V1T1=P2V2T2; V2=P1V1T1×T2P2;V2=750mm×1100.4mL×273K288K×760mm=1029.4mL. wt. of 1mL of H2=0.00009g∴ wt. of 1029.4mL of H2=0.00009×1029.4g=0.092646g. Weknow: wt. of metal wt. of H2 at NTP= Eq. wt. of metal Eq. wt. of H2(=1.008)∴ Eq. wt. of metal = wt. of metal wt. of H2 at NTP×1.008 =3.1×1.0080.092646=33.7 Ans.

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