A white sodium salt X, when warmed with dilute $H X_{2} SO X_{4}$ , gives a pungent gas that turns acidified dichromate paper green and a pale yellow turbidity, and no gas blackens lead acetate paper. A sample of X containing 0.050 mol of the anion is warmed with excess dilute acid and all the gas evolved is passed into excess acidified $KMnO X_{4}$ (the yellow solid is not attacked). The amount of $MnO X_{4} X^{-}$ reduced is

Answer: B

The observations identify X as

Na X_{2} S X_{2} O X_{3}

S X_{2} O X_{3} X^{2 -} + 2 H X^{+} SO X_{2} + S + H X_{2} O

gives 1 mol

SO X_{2}

per mole of anion, so 0.050 mol

SO X_{2}

. Permanganate oxidises it as

5 SO X_{2} + 2 MnO X_{4} X^{-} + 2 H X_{2} O 5 SO X_{4} X^{2 -} + 2 Mn X^{2 +} + 4 H X^{+}

. Hence

n (MnO X_{4} X^{-}) = 0.050 \times 2/5 = 0.020

mol.