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JEE Main levelNumerical answerDifficulty 2 of 3pb139

Potash alum, $KAl (SO X_{4}) X_{2} \cdot 12 H X_{2} O$ (9.48 g), is strongly ignited until only $Al X_{2} O X_{3}$ remains as the solid residue (K, S and water are lost or taken as other products). Use atomic masses K = 39, Al = 27, S = 32, O = 16, H = 1. The mass of $Al X_{2} O X_{3}$ obtained, in grams, to two decimal places, is

Numerical answer type, answer in g, tolerance 0.01.

Answer: 1.02 g

Molar mass of the alum

= 39 + 27 + 2 (32 + 64) + 12 (18) = 474 g/mol

. Moles

= 9.48/474 = 0.0200

. Each mole of alum gives 0.5 mol of

Al X_{2} O X_{3}

(M = 102 g/mol), so mass

= 0.0200 \times 0.5 \times 102 = 1.02 g

.

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