What is the freezing point ( in C ) of a 0.321 m aqueous solution of NaCl ? Enter your rounded answer with 3 decimal places.

A solution exhibits colligative properties since solute interferes with the interaction of the atoms which would change certain properties of the solution. It's the freezing point that is affected. Here's how we solve it:

ΔT = kf (m) i

where kf is the freezing point depression constant which is 1.86 for water

m= molality 

i= the van't hoff factor which is 2 for NaCl

ΔT = 1.86(0.321) 2

ΔT = 1.19412 °C

1.19412 °C = 0 - T

T = -1.19412 °C

Round your answer with 3 decimal places, which is -1.194°C

Hope this helps and stay safe during these crucial we will get through this!