Calculate the molality of a solution containing 5.3 g of anhydrous Na2CO3 in 400 g water .

Explanation

We are given the weight of solution = 214.2 g
Weight of water in solution `(w_(A))=214.2-34.2=180.0g`
Weight of sugar in solution `(w_(B))=34.2g`
We are to calculate the mole fraction of sugar `(x_(B))`
We know that `n_(B)=w_(B)/(M_(B))=((34.2g))/((342gmol^(-1)))=0.1mol,n_(A)=w_(A)/(M_(A))=(180g)/(18gmol^(-1))=10mol`
Using the relation,
`x_(B)=n_(B)/(n_(B)+n_(A))=0.1/(0.1+10.0)=0.1/(10.1)=0.0099`
Therefore, mole fraction of sugar is 0.0099.