Answer to Question #316536 in General Chemistry for miles

Question

1.      What will be the freezing point of an aqueous solution containing 80 g of NaCl? 

Answer 1

I don't think the question is complete. It's missing the volume of the solvent (probably water) so I'd just walk you through it.

∆T_F= K_F*m*i

i = van’t hoff factor of the solute.

m = molality of the solution.

K_f = Freezing point depression constant.

NaCl dissociates into two ions, so i= 2

Like I said earlier, I'd assume the solvent is water, so K_F = 1.86⁰C/m number of moles of NaCl = 80/(23+35.5) = 1.4 mol of NaCl

Let the volume of water (solvent) be V

Density = mass/ volume (density of water is 1kg/L)

1 = mass/V

Mass = V in kg

Molality = moles of NaCl/mass of solvent (water) in kg

= 1.4/ V

At this point, we have all our parameters and we can input them into the formula

Whatever you get is the CHANGE in freezing point and not the freezing point

To get the freezing point,

∆T_F = T_solution - T_water

T_solution = ∆T_{F +} T_water

note T_water = 0⁰C