Question

Correct expression among the following are: (\[\rho \]- density of solution; M- molarity of solution, \[{{M}_{2}}\]- molar mass of solute; m- molality; \[{{X}_{2}}\]- mole fraction of solute)
A. \[m=\dfrac{1000M}{1000\rho -M{{M}_{2}}}\]
B. \[{{X}_{2}}=\dfrac{m{{M}_{1}}}{1+m{{M}_{1}}}\]
C. \[{{X}_{2}}=\dfrac{M{{M}_{1}}}{M({{M}_{1}}-{{M}_{2}})+\rho }\]
D. \[m=\dfrac{\rho -M{{M}_{2}}}{M}\]

Answer 1

Hint: We know that the molarity is defined as the number of moles of solute which is present in per liters of the solution. Molality is defined as the number of moles of solute which are present in one kilogram of solvent. Mole fraction is defined as the unit of measurement in which the total moles of the substance in the solution is divided by the total moles of solution.

Complete step by step solution:
The formula for mole fraction is the following
\[{{X}_{2}}=\dfrac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}=\dfrac{{{n}_{2}}}{{{m}_{1}}/{{M}_{1}}+{{n}_{2}}}=\dfrac{{{M}_{1}}{{n}_{2}}}{{{m}_{1}}+{{M}_{1}}{{n}_{2}}}=\dfrac{{{M}_{1}}{{n}_{2}}}{\left( {{m}_{1}}+{{m}_{2}} \right)+{{M}_{1}}{{n}_{2}}-{{m}_{2}}}=\dfrac{{{M}_{1}}{{n}_{2}}}{V\rho +{{n}_{2}}\left( {{M}_{1}}+{{M}_{2}} \right)}\]
Here, \[{{n}_{2}}\] Is the number of moles of the substance let it be the solute
\[{{n}_{1}}\]Is the number of moles of solvent
\[\Rightarrow \dfrac{{{M}_{1}}\left( {{n}_{2}}/V \right)}{\rho +\left( \dfrac{{{n}_{2}}}{V} \right)\left( {{M}_{1}}+{{M}_{2}} \right)}=\dfrac{M{{M}_{1}}}{\rho +M\left( {{M}_{1}}-{{M}_{2}} \right)}\]
We can write the \[{{n}_{1}}\]as\[{{n}_{1}}=\dfrac{{{m}_{1}}}{{{M}_{1}}}\] here the \[{{m}_{1}}\] is referred as mass of the substance present in the solution and \[{{M}_{1}}\]is the molecular or atomic mass of the substance.
Also, \[m=\dfrac{1000M}{1000\rho -M\times {{M}_{2}}}\] Let, mass of solute be \[{{W}_{2}}\] ​ and solvent be \[{{W}_{1}}\]
Thus, $M=\dfrac{{{W}_{2}}}{{{M}_{2}}}\times \dfrac{100}{V}$ and $m=\dfrac{W}{{{M}_{2}}}\times \dfrac{1000}{{{W}_{1}}}$
Therefore, $\rho =\dfrac{W+{{W}_{2}}}{V}$ from $M$ and $\rho $ we can derive the value of $V,$
$\Rightarrow V=\dfrac{{{W}_{2}}\times 1000}{M{{M}_{2}}}$
So now we have, ${{W}_{1}}+{{W}_{2}}={{W}_{2}}+\left( \dfrac{1000{{W}_{2}}}{m{{M}_{2}}} \right)$ we get;
$\Rightarrow {{W}_{1}}+{{W}_{2}}={{W}_{2}}+\left( \dfrac{m{{M}_{2}}}{1000m}+\dfrac{1}{m} \right){{W}_{1}}$
From here we have get the value of $m$
$\therefore \dfrac{\rho }{M}=\dfrac{1}{m}+\dfrac{{{M}_{2}}}{1000}$
$\Rightarrow m=\dfrac{M\times 1000}{1000\rho -M{{M}_{2}}}$

So, the correct answer is Option A.

Note: Normality is defined as the molar concentration of either of the acid component or of the base component in the solution. The relationship between the two is \[normality\times equivalent\left( mass \right)=molarity\times molar\left( mass \right)\], The SI unit of molarity is mol/L and for normality is $eq/L.$