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» <html><body><p></p>Solution :`W_(2) = 2.56 g , W_(1) = 100g` <br/> `T = 319.692K , K_(b) = 2.42 K kg mol^(-1)` <br/> `Delta T_(b) = (319.692- 319.450)K = 0.242K` <br/> `M_(2) = (K_(b) xx W_(2) xx <a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)/(Delta T_(b) xx W_(1))= (2.42 xx 2.56 xx 1000)/(0.242 xx 100)` <br/> `M_(2) = 256g mol^(-1)` <br/> Molecular mass of sulphur in solution = `256g mol^(-1)` <br/> Atomic mass of one mole of sulphur atom = 32 <br/> No. of atoms in a molecule of sulphur `= (256)/(32) =8` <br/> Hence, molecular formula of sulphur is `S_(8)`. <br/> (ii) Consider a solution <a href="https://interviewquestions.tuteehub.com/tag/containing-2035611" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINING">CONTAINING</a> two components A and B <a href="https://interviewquestions.tuteehub.com/tag/whose-1455415" style="font-weight:bold;" target="_blank" title="Click to know more about WHOSE">WHOSE</a> 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. <br/> `x_(A) = (n_(A))/(n_(A) + n_(B)), x_(B) = (n_(B))/(n_(A) + n_(B))` <br/> `x_(A) + x_(B) = (n_(A))/(n_(A) + n_(B)) + (n_(B))/(n_(A) + n_(B)) =1`</body></html>


Discussion

No Comment Found

Related InterviewSolutions