One gram of an alloy of Al and Mg when treated with excess of dilute HCI forms MgCI_2, AICI_3 and hydrogen. The evolved hydrogen collected over Hg at 0°C has a volume of 1.20 litres at 0.92 atm pressure. Calculate the composition of the alloy. ("AI" = 27, "Mg" = 24)

One gram of an alloy of Al and Mg when treated with excess of dilute H

1.	One gram of an alloy of Al and Mg when treated with excess of dilute HCI forms MgCI_2, AICI_3 and hydrogen. The evolved hydrogen collected over Hg at 0°C has a volume of 1.20 litres at 0.92 atm pressure. Calculate the composition of the alloy. ("AI" = 27, "Mg" = 24)
Answer» Solution :Volume of `H_2` produced at NTP `= ( 1.2 xx 0.92)/( 273) xx (273)/(1)` `((p_1 V_1)/( T_1) = ( p_2 V_2)/( T_2) )` `= 1.104` LITRES. SINCE both `Al and Mg` produce `1.104` litres of `H_2`, equivalent of `Al+` equivalent of Mg `=`equivalent of hydrogen. `THEREFORE ("weight of Al")/( "eq. wt. of Al") + ("weight of Mg")/("eq. wt. of Mg")="eq. of hydrogen" = ("volume of hydrogen (NTP)")/("equivalent volume of hydrogen")`. Let the weight of `AI` be `x` g. `therefore (x)/( 27//3) + (1-x)/( 24//2) = (1.104)/( 11.2) ""(E_("AI")= (27)/(3) )` `(E_("Mg") = (24)/(2) )` ` x = 0.55` `(1-x) = 0.45`. Thus, weight of `"AI" = 0.55 g` and weight of `"Mg" = 0.45 g`.