Question

mole fraction of ethyl alcohol in a solution containning 36g of water and 46g of ethy alcohol

Answer 1

Mole fraction of ethyl alcohol is  $\frac{1}{3}$ .

Explanation:

Mole fraction of any solution is equal to mole of one component divided by total number of moles present in the solution.

                    $Mole fraction ( X) =\frac{mole of one component}{total moles of solution}$

We know that     $Mole (n)= \frac{Weight}{molecular weight}=\frac{w}{m.wt}$

Molecular weight of ethyl alcohol (A) = 46

Molecular weight of water (B) = 18

Given- weight of ethyl alcohol = 46 g

weight of water = 36 g

Suppose mole fraction of ethyl alcohol  = $X_{A}$

mole fraction of water = $X_{B}$

          $X_{A}=\frac{n_{A} }{(n_{A}+n_{B})}=\frac{\frac{weight of A}{m.wt of A} }{\frac{weight of A}{m.wt of A}+\frac{weight of B}{m.wt of B} } =\frac{\frac{46}{46} }{\frac{46}{46}+\frac{36}{18} } =\frac{1}{1+2}=\frac{1}{3}$  

          $X_{B}=\frac{n_{B} }{(n_{A}+n_{B})}=\frac{\frac{weight of B}{m.wt of B} }{\frac{weight of A}{m.wt of A}+\frac{weight of B}{m.wt of B} } =\frac{\frac{36}{18} }{\frac{46}{46}+\frac{36}{18} } =\frac{2}{1+2}=\frac{2}{3}$

So, the mole fraction of ethyl alcohol and water should be 1/3 and 2/3 respectively.