What's the relation between molarity and molality?

I am trying to derive the relation between the [molarity](https://en.wikipedia.org/wiki/Molar_concentration) and [molality](https://en.wikipedia.org/wiki/Molality) of a solution



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



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> Let $V\_{\text{sol}}= $ volume of solution
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> Molarity (M) $= n/V\_{\text{sol}}$
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* 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
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> molality $(m) = n/W$
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> 
> 


* 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 ($M$) = $n$/$V\_{\text{sol}}$
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> 
> Molality ($m$) = $n$/$W$
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> 
> 


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 $V\_{\text{sol}}$.


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