The density of `0.06 M` solution of `Kl` in water is `1.006 g mL^(-1)`. Determine the molality of this solution `(K = 39, I = 127 "amu")`

Explanation

`Mw of Kl = 39 + 127 = 166 g`
`d = M ((Mw_(2))/(1000) + (1)/(m))`
`1.006 = 0.006 ((166)/(1000) + (1)/(m))`
Solve for `m`,
`m = 0.06024`
Alternate method:
`M = (% "by Weight" xx 10 xx d)/(Mw_(2))`
`0.06 = (%" "by Weight" xx 10 xx 1.006)/(166)`
`:. "by weight" = (166 xx 0.06)/(10 xx 1.066) = 0.99%`
Weight of solute `= 0.99 g`.
Weight of solution `= 100 g`
Weight of solvent `= 100 - 0.99 = 99.01`
`m = (w_(2) xx 1000)/(Mw_(2) xx w_(1)) = (0.99 xx 1000)/(166 xx 99) = 0.06024`