The relation between molarity `(C )` and molality `(m)` is given by )d= density of solution, `M` =molar mass of solute):

Options

A. `m=(1000 CM)/(1000 C-d)`
B. `m=(1000 C)/(1000 d-CM)`
C. `m=(Cd)/(1000d-CM)`
D. `m=(CMd)/(1000-dM)`

Answer

2

Explanation

`Molality (C )=n_("solute")/V_(L)`
`"Molality "(m)=n_("solute")/(kg_("solvent"))`
Thus, `m/C=V_(L)/(Kg_("solvent"))`
`Kg_("solvent")= mass of solution in kg - mass of solute in Kg`
`=[(("density of"),("solution in Kg "L^(-1)))(("volume of"),("soln in "L))]`
`-[(n_("solute")xx" Molar mass of solute"),("in kg "mol^(-1))]`
`=[d(kg L^(-1))V_(L)]-[C.V_(L)xxM/1000]`
where M is the molar mass of solute in g and 1000 is the conversion factor to get molar mass in kg.
Finally
`Kg_("solvent")=V_(L)[d-(CM)/1000]`
Substituting this result in Equlation (1), we get
`m/C=V_(L)/(V_(L)[d-(CM)/1000])=1/([d-(CM)/1000])`
`m=C/(d-(CM)/1000)`
`=C/((1000 d-CM)/1000)`
`m=(1000 C)/(1000 d-CM)`
where m= molality in mol `Kg^(-1)`
C= molarity in mol `L^(-1)`
d= density of solution in `Kg L^(-1)`
M= molar mass of solute in `g mol^(-1)`
If molar mass of solute is expressed as `Kg mol^(-1)` then 1000 will be replaced by 1, that is,
`m=C/(d-CM)`