1, calculate the vapor pressure of solution A at 298K Ans: 19.0 torr 2, calculat

Question

1, calculate the vapor pressure of solution A at 298K Ans: 19.0 torr 2, calculate the freezing point of solution B Ans: -19 Celcius 3, calculate the boiling point of solution C Ans: 108 Celcius 4, what would the vapor pressure of pure water be at 350K? Ans: 296 torr
I DONT KNOW HOW TO GET TO THE SOLUTION SOLUTIONS A) 5.00 m FeCis(aq) c) 5.00 m MgCl2(a) B) 5.00 m NaCl(aa) D) 5.00 m CoH120s (glucose)(aa) (aq) WATER CONSTANTS K, = 1.86 °C/m AHyan40.7 kJ/mol Kb = 0.512 °C/m Vapor pressure 0298K-25,8 torr

Explanation / Answer

Molality of FeCl3=5

Moles of solute/kg ofwater= 5

Basis : 1 kg of water= 1000gms of water

Moles of water =1000/18=55.56 , total moles of solution= 55.56+5= 60.56

Mole fraction water= 55.56/60.56=0.917

Vapor pressure of water in solution= mole fraction* pure component vapor pressure =0.917*25.8=23.65 torr

b)

Freezing point depression= i*kf*m

I= Van’t Hoff factor= 2 for NaCl, Kf= 1,86deg.c/m for water and m= 5m

Freezing point depression =2*1.86*5=18.6

Freezing point of water- freezing point of solution =18,6

Freezing point ofsolution =0-18.6= -18.6 deg.c

c)

Boiling point elevation= i*kb*m

I= 3 for MgCl2 ( since MgCl2 spilts into Mg+2 and 2 Cl- ion)

Kb=0.512 and m=5

Boiling point elevation= 3*0.512*5= 7.68

Boiling point of solution- boiling point of water= 7.68

Boiling point of solution =100+7.68= 107.68 deg.c

d)

From Classius-Clyperon equation

Ln(P2/P1)=(delH/R)*(1/T1-1/T2)

Given T1= 298K T2= 350 K, P1= 25.8 Torr

delH= latent heat of vaporization of water =40.7 Kj/mol= 40.7*1000 joules/mol

R= 8.314 joules/mole

Ln(P2/25.8)= (40.7*1000/8.314)*(1/298-1/350)=2.44

P2= 25.8*11.48=296.184 Torr

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