Question

what is the boiling point of a 0.743m aqueous solution of KCl? (report amount to three decimal points)

Answer 1

ΔTb = Kb · bBPermalink Submitted by kingchemist on Sun, 2014-06-01 09:41ΔTb = Kb · bBWhere Kb = ebullioscopic constant for water and bB= molality of solution x van' Hoff factor which is 2 foer K+Cl- as there are two particles per formula.or = 2 x Kb x 0.743

Answer 2

Answer: The boiling point of a 0.743m aqueous solution of KCl is
`100.760^o C`.

Step-by-step explanation:

`\Delta T_b=i* K_b* m`

`\Delta T_b` = change in boiling point

i= Vant hoff factor =is the ratio of observed colligative property to the calculated colligative property.

`KCl\rightarrow K^(+)++Cl^(-)`

`K_b` = boiling point constant for water = 0.512 °C kg/mol

m= molality

`\Delta T_b=T_b-T_b^0=(T_b-100)^0C`

`(T_b-100)^0C=2* 0.512* 0.743}[/tex ]</p><p>[tex]T_b=100.760^C`

The boiling point of the solution is
`100.760^o C`.