1.00 g of `BaCl_2` is treated with excess of aqueous `AgNO_3` and all chlorine is recovered as 1.38 g of AgCl. What is the atomic weight of Ba? (`Cl^(-) 35.5, Ag^(-) 108)`

Options

A. 137
B. 172.5
C. 33
D. 68.5

Answer

A

Explanation

` underset(1 g) ( BaCl_2) + 2 Ag NO_3 to underset(1.38 g ) ( 2 AgCl ) + Ba (NO_3)_2`
All chlorine atoms are recovered as AgCl hence number of moles of chlorine on both sides must be equal. Thus, 2 `xx` number of moles of `BaCl_2 = 1 xx` number of moles of AgCl
` 2 xx (1)/( ( x + 35.5 xx 2 ) = 1xx ( 1.38 )/((108 +35.5 ) ) implies x ~~ 37`
(Say atomic mass of Ba = x)