Question

What is the `"molality"` of an aqueous solution of `KCl`, if `"molarity"=1*mol*L^-1`, and `rho_"solution"=1.04*g*mL^-1`?

Answer 1

`"What will be molality of a KCl solution..........."` `"if MOLARITY"`
`"is 1"*mol*L^-1,` `"and"` `rho=1.04*g*mL^-1`.

Explanation:

We want the ratio.......

`"molality"="mole of solutes"/"kilograms of solvent"`.

Now in `1*L` of solution there are `1*mol` salt........, i.e. a mass of `74.55*g` salt.............. But, by specification, this SOLUTION had a mass of `1*Lxx1000*mL*L^-1xx1.04*g*mL^-1=1040*g`.

Note that I made the assumption that you meant to write `rho=1.04*g*mL^-1` NOT `1.04*g*L^-1`.

Since masses are certainly additive, given these data, we can now address the quotient...........

`"molality"="mole of solutes"/"kilograms of solvent"`

`=(1*mol_"KCl")/((1040-74.55)*gxx10^-3*kg*g^-1)=(1*mol_"KCl")/(0.9645*kg)`

`=1.037*mol*kg^-1=1.04*mol*kg^-1`....to three sigs.........