Question

1. The mole fraction of an aqueous solution of calcium sulfate is 0.225. Calculate the molarity (in mol/L) of the calcium sulfate solution, if the density of the solution is 1.05 g mL^-1.

2. Determine the mole fraction of potassium acetate in a 7.19 M aqueous solution of potassium acetate. The density of the solution is 1.30 g mL^-1.

Answer 1

1)density of the solution=1.05 g/ml

Molar mass=[CaSO4=40+32+4*16=136 g/mol]

Molar mass of H2O=2*1+16=18g/mol

X(CaSO4)=Xc=0.225(mole fraction)

Let nc=moles of calcium sulfate and nw=moles of water

Xc=nc/(nc+nw)……..by definition

1/Xc=(nc+nw)/nc

Or, 1/Xc=1+nw/nc

Or, 1/Xc- 1=nw/nc

Or,(1/0.225)-1=nw/nc

1/Xc=(nc+nw)/nc

Or, 1/Xc=1+nw/nc

Or, 1/Xc- 1=nw/nc

Or,(1/0.225)-1=nw/nc

Or,3.44=nw/nc

Or,nw/nc= (mass of w/molar mass of w)/ (mass of c/molar mass of c)

=(mass of w/mass of c)*Mc/Mw

  =(mass of w/mass of c)*(136 g/mol/18g/mol)

   =(mass of w/mass of c)*7.55

3.44= (mass of w/mass of c)*7.55

Or, 3.44/7.55= (mass of w/mass of c)

Or, 3.44/7.55= (mass of w/mass of c)

0.45+1=(mass of w/mass of c)+1

1.45=mass of w+mass of c/mass of c

1/1.45= mass of c/ (mass of C+mass of w)=mass of c /total mass=mass fraction of C

0.689=mass fraction of C

Mass fraction of C=0.689

Given mass in 1L of solution=1050g

Mass of C=0.689*1050g=723.45g

Moles of C=723.45g/molar mass==723.45g/136g/mol=5.32 moles/L=5.32M

2) molarity of potassium acetate =7.19 M =7.19mol/L

Mass of potassium acetate in 1L=7.19 moles*molar mass of potassium acetate

[CH3COOK=C2H3O2K=2*12+3*1+2*16+40=24+3+32+40=99 g/.mol]

Mass of potassium acetate in 1L=7.19 moles*99g/mol=711.81 g

density of the solution = 1.30 g mL-1

total mass in 1L solution=1300 g

mass of water=1300-711.81=588.19 g

moles of water=588.19g/18g/mol=32.68 moles

X(potassium acetate)=mole fraction=n(potassium acetate)/n(total)=7.19/(7.19+32.68)=7.19/39.87=0.18

X(potassium acetate)=mole fraction=0.18