what is the boiling point of a 0.321 m aqueous solution of NaCl. Enter your rounded answer with 3 decimal places.

∆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 = 100.329ºC

So Our Answer is 100.329ºC

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