Calculate the molarity of `KCl` solution prepared by dissolving `7.45 g` of `KCl` in `500 mL` of the solution. `(d_(sol) = 1.2 g mL^(-1))`

Explanation

Molar mass of `KCl = (39 + 35.5) g = 74.5 g`
` W_(2) 7.45 g, Mw_(2) = 74.5 g`
`V_(sol) = 500 mL`
`d_(sol) = 1.2 g mL^(-1)`
`m = (W_(2) xx 1000)/(Mw_(2) xx W_(1))`
In the above relation, `W_(1)` is known, so find `W_(1)`.
`W_(1) = (W_(sol) - W_(2)) g = (V_(sol) xx d_(sol) - W_(2)) g`
`(500 xx 1.2 - 74.5) g = 525.5 g`
`:. m = (7.45 xx 1000)/(74.5 xx 525.5) = 0.19 mol kg^(-1) = 0.19 m`