2. What is the freezing point of an aqueous solution containing 69.5g NaNO3 (MM=85.00g/mol) dissolved in 175g water? Kf=1.86°C/m for water and assume "ideal" behavior.
10. An RV antifreeze is an aqueous solution containing 5.68 M proplyene glycol (MM=76.09 g/mol). If the density of the antifreeze is 1.080 g/mL, what is the molality of the solution?
2)
Lets calculate molality first
Molar mass of NaNO3,
MM = 1*MM(Na) + 1*MM(N) + 3*MM(O)
= 1*22.99 + 1*14.01 + 3*16.0
= 85 g/mol
mass(NaNO3)= 69.5 g
use:
number of mol of NaNO3,
n = mass of NaNO3/molar mass of NaNO3
=(69.5 g)/(85 g/mol)
= 0.8176 mol
m(solvent)= 175 g
= 0.175 kg
use:
Molality,
m = number of mol / mass of solvent in Kg
=(0.8176 mol)/(0.175 Kg)
= 4.672 molal
NaNO3 dissociates into Na+ and NO3-.
So, i=2
lets now calculate ΔTf
ΔTf = i*Kf*m
= 2*1.86*4.6723
= 17.4 oC
This is decrease in freezing point
freezing point of pure liquid = 0.0 oC
So, new freezing point = 0 - 17.4
= -17.4 oC
Answer: -17.4 oC
I am allowed to answer only 1 question at a time
2. What is the freezing point of an aqueous solution containing 69.5g NaNO3 (MM=85.00g/mol) dissolved in...
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