A 0.500 g sample of pure anhydrous Na2SO4 is dissolved in water, acidified with dilute HCl and treated with excess BaCl2 solution. Taking Na 23, S 32, O 16 and Ba 137, calculate the mass of dry BaSO4 obtained, in grams, correct to two decimal places.

JEE Main levelNumerical answerDifficulty 2 of 3qa067

A $0.500 g$ sample of pure anhydrous $Na X_{2} SO X_{4}$ is dissolved in water, acidified with dilute HCl and treated with excess $BaCl X_{2}$ solution. Taking Na 23, S 32, O 16 and Ba 137, calculate the mass of dry $BaSO X_{4}$ obtained, in grams, correct to two decimal places.

Numerical answer type, answer in g, tolerance 0.01.

M (Na X_{2} SO X_{4}) = 142

and

M (BaSO X_{4}) = 233 g mo l^{- 1}

, and

Ba X^{2 +} + SO X_{4} X^{2 -} BaSO X_{4}

is 1:1. Moles of sulphate

= 0.500/142 = 3.521 \times 1 0^{- 3}

. Mass

= 3.521 \times 1 0^{- 3} \times 233 = 0.8204 g \approx 0.82 g