# What's the relation between molarity and molality?

I am trying to derive the relation between the molarity and molality of a solution

Mole concept $$\displaystyle n = \frac{M'}{M''}$$

• Here, ($$n$$) stands for number of moles
• $$\ce{(M^{'})}$$ stands for mass of substance in gram
• $$\ce{(M^{''})}$$ stands for molecular/atomic mass of substance

Let $$V_{\text{sol}}=$$ volume of solution

Molarity (M) $$= n/V_{\text{sol}}$$

• The above equation of molarity ($$M$$) shows the concentration of solution of substance, i.e. ($$M$$) is the ratio of number of moles ($$n$$) with respect to it's volume ($$V_{\text{sol}}$$)

let $$W$$ = mass of solvent in kilogram

molality $$(m) = n/W$$

• The above equation of molality ($$m$$) shows the concentration of solution of substance
• i.e. ($$m$$) is the ratio of number of moles ($$n$$) with respect to mass of it's solvent ($$W$$)

What is the relation between molarity ($$M$$) and molality ($$m$$)?

Molarity ($$M$$) = $$n$$/$$V_{\text{sol}}$$

Molality ($$m$$) = $$n$$/$$W$$

I'm confused on how to relate the $$W$$ with $$V_{\text{sol}}$$.

$W$ and $V_{\text{sol}}$ can be related via the density of solution. We have, mass of solution = mass of solute + mass of solvent.
Hence, $$M_{\text{sol}} = W + M’$$
Let us assume the density of the solution to be $\rho$. Since, density is equal to mass over volume, we have, $$\rho= \frac{M_{\text{sol}}}{V_{\text{sol}}} = \frac{W + M’}{V_{\text{sol}}}$$ Thus, you have relation between $W$ and $V_{\text{sol}}$.