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

Jun 3, 2017

$\text{What will be molality of a KCl solution...........}$ $\text{if MOLARITY}$
$\text{is 1} \cdot m o l \cdot {L}^{-} 1 ,$ $\text{and}$ $\rho = 1.04 \cdot g \cdot m {L}^{-} 1$.

#### Explanation:

We want the ratio.......

$\text{molality"="mole of solutes"/"kilograms of solvent}$.

Now in $1 \cdot L$ of solution there are $1 \cdot m o l$ salt........, i.e. a mass of $74.55 \cdot g$ salt.............. But, by specification, this SOLUTION had a mass of $1 \cdot L \times 1000 \cdot m L \cdot {L}^{-} 1 \times 1.04 \cdot g \cdot m {L}^{-} 1 = 1040 \cdot g$.

Note that I made the assumption that you meant to write $\rho = 1.04 \cdot g \cdot m {L}^{-} 1$ NOT $1.04 \cdot g \cdot {L}^{-} 1$.

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

$\text{molality"="mole of solutes"/"kilograms of solvent}$

$= \frac{1 \cdot m o {l}_{\text{KCl")/((1040-74.55)*gxx10^-3*kg*g^-1)=(1*mol_"KCl}}}{0.9645 \cdot k g}$

$= 1.037 \cdot m o l \cdot k {g}^{-} 1 = 1.04 \cdot m o l \cdot k {g}^{-} 1$....to three sigs.........