2.56g of Sulphur is dissolved in 100g of carbon disulphide. The solution boils at 319.692 K. What is the molecular formula of Sulphur in solution. The boiling point of CS_(2) is 319.450K. Given that K_(b) for CS_(2) = 2.42 Kkg mol^(-1) (ii) Show that the sum of mole fraction of a solution is equal to one.

2.56g of Sulphur is dissolved in 100g of carbon disulphide. The soluti

1.	2.56g of Sulphur is dissolved in 100g of carbon disulphide. The solution boils at 319.692 K. What is the molecular formula of Sulphur in solution. The boiling point of CS_(2) is 319.450K. Given that K_(b) for CS_(2) = 2.42 Kkg mol^(-1) (ii) Show that the sum of mole fraction of a solution is equal to one.
Answer» Solution :`W_(2) = 2.56 g , W_(1) = 100g` `T = 319.692K , K_(b) = 2.42 K kg mol^(-1)` `Delta T_(b) = (319.692- 319.450)K = 0.242K` `M_(2) = (K_(b) xx W_(2) xx 1000)/(Delta T_(b) xx W_(1))= (2.42 xx 2.56 xx 1000)/(0.242 xx 100)` `M_(2) = 256g mol^(-1)` Molecular mass of sulphur in solution = `256g mol^(-1)` Atomic mass of one mole of sulphur atom = 32 No. of atoms in a molecule of sulphur `= (256)/(32) =8` Hence, molecular formula of sulphur is `S_(8)`. (ii) Consider a solution CONTAINING two components A and B WHOSE mole fractions are `x_(A) and x_(B)` respectively. Let the number of moles of two components A and B are `n_(A) and n_(B)` respectively. `x_(A) = (n_(A))/(n_(A) + n_(B)), x_(B) = (n_(B))/(n_(A) + n_(B))` `x_(A) + x_(B) = (n_(A))/(n_(A) + n_(B)) + (n_(B))/(n_(A) + n_(B)) =1`