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JEE Advanced levelSingle correct optionDifficulty 3 of 3qc051

A colourless salt solution turns turbid white on dilution with water. With $H X_{2} S$ in dilute $HCl$ it gives a dark brown precipitate that is insoluble in yellow ammonium sulphide. Taking the cation to be $Bi X^{3 +}$ from $5.00 g$ of $BiCl X_{3}$ (Bi 209, Cl 35.5, S 32; $2 BiCl X_{3} + 3 H X_{2} S Bi X_{2} S X_{3} + 6 HCl$ ), the mass of the sulphide formed on complete precipitation is

A3.31 g
B4.07 g
C2.04 g
D8.15 g

Answer: B

M (BiCl X_{3}) = 209 + 3 \times 35.5 = 315.5 g mo l^{- 1}

, so

n = 5.00/315.5 = 0.01585 mol

. The equation gives

n (Bi X_{2} S X_{3}) = n /2 = 0.00792 mol

, and

M (Bi X_{2} S X_{3}) = 2 \times 209 + 96 = 514

. Mass

= 0.00792 \times 514 = 4.07 g

.

Practice Qualitative Analysis: Cations More on Group I and group II cations (Pb, Cu, Bi, Cd, As, Sb, Sn)

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