Respuesta :
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
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:
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 =