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The specific gravity of a salt solution ...

The specific gravity of a salt solution is 1.025. If `V mL` of water is added to `1 L` of this solution to make its density `1.02 g mL^(-1)`, what value of `V` in `mL` approxmately?

Text Solution

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`V_("New") = (V + 1000) mL`
Number of moles of salt remains same.
`:. (V + 1000) xx 10.2 = 1.025 xx 1000`
`V = 4.9 mL = 5 mL`
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