When dissolved in dilute sulphuric acid, 0.275 g of a metal evolved 119.7 mL of hydrogen at 20°C and 763 mm pressure. What is the equivalent mass of the metal ?

When dissolved in dilute sulphuric acid, 0.275 g of a metal evolved 11

1.	When dissolved in dilute sulphuric acid, 0.275 g of a metal evolved 119.7 mL of hydrogen at 20°C and 763 mm pressure. What is the equivalent mass of the metal ?
Answer» <P> Solution :In the PRESENT case, `P_1 = 763 mm, V_(1) = 119.7 ML, T_2 = 20°C = 293` K, MASS of metal = 0.275 g If the volume of hydrogen at N.T.P is `V_2`, we have from gas equation `(P_(1)V_(1))/T_(1) = (P_(2)V_(2))/T_(2)` or `(763 xx 119.7)/293 = (760 xx V_(2))/273` `therefore V_(2)=(793 xx 119.7 xx 273)/(760 xx 293) = 111.97` mL Mass of 111.97 mL of `H_2` at S.T.P. `=2/(22400) xx 111.97 = 0.011` g Hence, Equivalent weight of the metal `=("Mass of the metal")/("Mass of "H_(2) "evolved at S.T.P") = 0.275/0.011 = 25`