The amount of `H_2 S` required to precipitate 1.69 g BaS from `BaCl_2` solution is (Atomic weight Ba=137, S = 32 and H = 1) :

What is the amount of BaSO_4 Formed when 0.02 mole of BaCl_2 solution is treated with excess of Na_2SO_4 Solution,(Ba =137 , S= 32 , Na =23 )

Which one among the following is the equivalent weight of sulphuric acid ? (Atomic weight : H=1, S=32, O=16)

The equivalent weight of Ba(OH)_(2) is (given, atomic weight of Ba is 137.3)

0.32 g of an organic compound gave 0.233 g of BaSO_(4) . Determine the percentage of sulphur in the compound (Atomic mass of Ba = 137, S = 32, O = 16 )

0.32 g of an organic compound gave 0.233 g of BaSO_(4) . Determine the percentage of sulphur in the compound (Atomic mass of Ba = 137, S = 32, O = 16 )

A metal M forms the sulphate M_2(SO_4)_3 0.596 gm of the sample reacts with excess BaCl_2 to give, 1.22 gm BaSO_4 . What is the atomic weight of the M ? (S = 32, Ba + 137.3)

0.316 g of an organic compound, after heating with fuming nitric acid and barium nitrate crystals in a sealed tube gave 0.466 g of the precipitate of barium sulphate. Determine the percentage of sulphur in the compound. (Atomic masses : Ba = 137, S = 32, O = 16, C = 12, H = 1).