Question

Calculate a) molarity b)molality c) mole fraction of $KI$ if the density of 20% (mass by mass) aqueous $KI$ is 1.202$gm{{L}^{-1}}$ .

Answer 1

Hint: The answer to this question includes the basic chemistry formulas of molarity, molality and mole fraction given by molarity =$\dfrac{{{n}_{solute}}\times 1000}{volume}$, molality= $\dfrac{{{n}_{solute}}\times 1000}{{{M}_{KI}}}$ and mole fraction=${{\chi }_{KI}}=\dfrac{{{n}_{KI}}}{{{n}_{KI}}+{{n}_{{{H}_{2}}O}}}$

Complete answer:
We know from the basic concepts of chemistry that molarity of the solution is the ratio of number of moles of solute present in one litre of solution. From this definition, we can write the formula as,
a) Molarity= $\dfrac{{{n}_{solute}}\times 1000}{volume}$ …..(1)
Now, 20% mass by mass means 20g of $KI$is present in 100g of solution which is given by,
${{W}_{KI}}=20g$ and ${{W}_{{{H}_{2}}O}}=100-20=80g$ ……(2)
Since, in the question density is given we can write it as,
Density=$\dfrac{mass}{volume}$
Rearranging the above equation we have, Volume=$\dfrac{mass}{density}$
Substituting this in the equation number (1)
Molarity=$\dfrac{{{n}_{solute}}\times 1000}{{{\left( \dfrac{mass}{density} \right)}_{solution}}}$
Substituting values from equation (2) and that of density value, we get
Molarity=$\dfrac{20\times 1.202\times 1000}{166\times 100}$ [Since, molar mass of $KI$=166 g]
$\Rightarrow $ Molarity =1.448 M
b)Next we shall calculate molality of solution which is defined as ratio of number of moles of solute in 1kg of solvent which is given as,
Molality= $\dfrac{{{n}_{solute}}\times 1000}{{{M}_{KI}}}$
\[\Rightarrow \] Molality = $\dfrac{20\times 100}{166\times 80}$ [since weight of solvent=80g]
Therefore, molality=1.5 mol/kg
c) Mole fraction is defined as Ratio of one constituent to the total number of all constituents in a mixture. Therefore we can write as,
Mole fraction of $KI$,${{\chi }_{KI}}=\dfrac{{{n}_{KI}}}{{{n}_{KI}}+{{n}_{{{H}_{2}}O}}}$
\[\Rightarrow {{\chi }_{KI}}=\dfrac{\left( {20}/{166}\; \right)}{\left( {20}/{166}\; \right)+\left( {80}/{18}\; \right)}\]= $\dfrac{0.12}{0.12+4.44}$=0.0263
Thus, mole fraction of $KI$= 0.0263
Therefore, we have a) molarity of solution=1.448M , molality of solution = 1.5mol/kg and mole fraction = 0.0263

Note:
Most of the times students go wrong in writing units for the mole fraction same as that of molarity or molality. You should keep in mind the fact that mole fraction do not have any units since moth the units gets cancelled and it is just a numerical.