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 $100.760^o C$.

Explanation:

$\Delta T_b=i\times K_b\times m$

$\Delta T_b$ = 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\times 0.512\times 0.743}[/tex ]

[tex]T_b=100.760^C$

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