Question

What is the boiling point of a 0.321m aqueous solution of nacI

Answer 1

This might not make sense but just watch
NaCl - Na(+) +Cl(-) (two ions)

(0.321m NaCl) x(2 mol ions/1 mol NaCl) =
=0.642m ions

(0.642m) x (0.512°C/m) = 0.329°C change
100°C + 0.329°C = 100.33°C.
Good luck

Answer 2

Answer:

100.329ºC

Explanation:

∆T = imK

∆T = Change in boiling point (B.P.)

i = van't Hoff factor = 2 for NaCl (Na+ and Cl2_

m = molality = 0.321 m

k = boiling point constant for water = 0.512 deg/m

∆T = (2)(0.321)(0.512) = 0.329 degrees

Since the normal B.P. for water is 100ºC, the new boiling point of this solution is 100 + 0.329 =