Question

Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Answer 1

This can be solved using 2 formulae: ${{X}_{ethanol}}=\dfrac{{{n}_{ethanol}}}{{{n}_{ethanol}}+{{n}_{water}}}$ and $Molarity(M)=\dfrac{{{n}_{solute}}}{{{V}_{solution}}}$
First calculate the number of moles of water by taking given mass as 1000g.

Complete step by step solution:
Given that density of water$=1kg/{{m}^{3}}$
Therefore, water is approximately equal to 1Litre.
Molar mass of ${{H}_{2}}O=2\times 1+1\times 16=18g/mol$ .
The number of moles in 1L of water$=\dfrac{\text{given mass of water}}{\text{molar mass of water}}=\dfrac{1000}{18}=55.55moles$

& \text{Mole fraction of ethanol=}\dfrac{\text{no}\text{. of moles of ethanol}}{\text{total no}\text{. of moles of the solution}} \\\ & {{X}_{ethanol}}=\dfrac{{{n}_{ethanol}}}{{{n}_{ethanol}}+{{n}_{water}}} \\\ \end{aligned}$$ ${{X}_{ethanol}}=0.04$ (Given in the question) Substituting the values, $\begin{aligned} & 0.04=\dfrac{{{n}_{ethanol}}}{{{n}_{ethanol}}+55.55} \\\ & {{n}_{ethanol}}=2.31moles \\\ \end{aligned}$ 2.31 moles are present in 1L of solution. We already know that Molarity is defined as the number of moles of a solute added in a solution per liter of solution. $$Molarity(M)=\dfrac{{{n}_{solute}}}{{{V}_{solution}}}$$ $$M=\dfrac{2.31moles}{1L}=2.31M$$ **Therefore, the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 comes out to be 2.31M.** **Note:** Molarity should not be confused with molality. Molality(m) is defined as the number of moles of a solute per kilogram of a solvent whereas Molarity(M) is defined as the number of moles of a solute added in a solution per liter of solution.