1.

5g of `K_(2)SO_(4)` are dissolved in 250mL of solution. How many mL of this solution should be used so that 1.2 of `BaSO_(4)` may be precipated from `BaCl_(2)` solution?

Answer» The desired equation is
`BaCl_(2)+underset(=174g)underset(2xx39+32+64)(K_(2)SO_(4)) to underset(=223g)underset(137+32+64)(BaSO_(4))+2KCl`
`233g of BaSO_(4)` obtained from 174g of `K_(2)SO_(4)`
1.2 of `BaSO_(4)` will be obtained from `(174)/(233)xx1.2`
`=0.8961"g of "K_(2)SO_(4)`
So, 0.891g of `K_(2)SO_(4)` will be present in `(250)/(5)xx0.8961`
`=44.8"mL of solution"`


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