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JEE Main levelNumerical answerDifficulty 2 of 3qc033

$75.0 mL$ of $0.0860 M$ $CuSO X_{4}$ solution, acidified with dilute $HCl$ , is treated with excess $H X_{2} S$ . Assuming complete precipitation as $CuS$ (molar mass $95.5 g mo l^{- 1}$ from Cu 63.5, S 32), calculate the mass of $CuS$ formed in grams, correct to two decimal places.

Numerical answer type, answer in g, tolerance 0.01.

Answer: 0.62 g

n (Cu X^{2 +}) = 0.0860 \times 0.0750 = 0.00645 mol

. Since

Cu X^{2 +} + H X_{2} S CuS + 2 H X^{+}

is 1:1,

n (CuS) = 0.00645 mol

. Mass

= 0.00645 \times 95.5 = 0.6160 g \approx 0.62 g

.

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