Discussion Forum

The mole fraction of a solute in a solution is 0.1, At 298K , molarity of this solution is the same as its molality . Density of this solution at 298K is 2.0 g cm^-3 . The ratio of the molecular weights of the solute and solvent,  $(\frac{m_{solute}}{m_{solvent}})$ is.......

Answer:

Explanation:

 Moles of solute , $n_{1}=\frac{w_{1}}{m_{1}}$ ; Moles of solvent ,   $n_{2}=\frac{w_{2}}{m_{2}}$ 

$x_{1}$ (solute)=0.1 and   $x_{2}$ ( solvent)=0.9

$\therefore$    $\frac{x_{1}}{x_{2}}=\frac{n_{1}}{n_{2}}=\frac{w_{1}}{m_{1}}.\frac{m_{2}}{w_{2}}=\frac{1}{9}$

$Molarity=\frac{Solute(moles)}{Volume(L)}$

$=\frac{W_{1}\times 1000\times2}{m_{1}(w_{1}+w_{2})}$

 Note volume= Total mass of solution /  Density

  $=(\frac{w_{1}+w_{2}}{2})mL$

  $Molality=\frac{Solute(moles)}{Solvent(kg)}=\frac{w_{1}\times1000}{m_{1}\times w_{2}}$

 Given, Molarity=Molality

 hence,   $\frac{2000w_{1}}{m_{1}(w_{1}+w_{2})}=\frac{1000 w_{1}}{m_{1} w_{2}}$

 $\therefore$                 $\frac{w_{2}}{w_{1}+w_{2}}=\frac{1}{2}$

$\Rightarrow$    $w_{1}=w_{2}=1$

$\therefore$    $\frac{w_{1}m_{2}}{m_{1}w_{2}}=\frac{1}{9}$

$\Rightarrow$    $\frac{m_{1}(solute)}{m_{2}(solvent)}=9$