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Q.

A 10.00 g sample of a soluble barium salt is treated with an excess of sodium sulphate to precipitate $11 .21 g BaSO (M = 233 .4)$ . Which barium salt is it?

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a

${BaCl}_{2} (M = 208 .2)$

b

$Ba {(O_{2} CH)}_{2} (M = 227 .3)$

c

$Ba {({NO}_{3})}_{2} (M = 261 .3)$

d

${BaBr}_{2} (M = 297 .1)$

answer is A.

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Detailed Solution

No. of mole is found as,

$n = \frac{m}{M}$

Since By principle of Atom conservation

Mass is same

$M = \frac{m}{n} M_{1} = M_{2} \frac{m_{1}}{n_{1}} = \frac{m_{2}}{n_{2}}$

$\frac{10}{M} = \frac{11 .21}{233 .4}$

M = 208.2 ( ${BaCl}_{2}$ )

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A 10.00 g sample of a soluble barium salt is treated with an excess of sodium sulphate to precipitate 11.21g BaSO(M=233.4). Which barium salt is it?