10 mL of a solution of K2SO4, on reaction with BaCl2, gave 0.233 g of white ppt. The concentration of K2SO4 solution is: (Given that the molecular weight of Ba=137,S=32,O=16,K=39)
A
17.4gL−1
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B
34.8gL−1
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C
8.7gL−1
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D
4.4gL−1
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Solution
The correct option is A17.4gL−1 The balanced chemical equation is K2SO4+BaCl2→BaSO4+2KCl 0.233 g of BaSO4 corresponds to 0.233233=0.001 moles. They also corresponds to 0.001 moles of K2SO4. 0.001 moles in 10 ml will be equal to 0.1 moles in 1 L. It is equual to 174g/mol×0.1M=17.4g/L of K2SO4.