A solution is prepared from 4.5701 g of magnesium chloride and 43.238 g of water. The vapor pressure of water above this solution is 0.3624 atm at 348.0 K. The vapor pressure of pure water at this temperature is 0.3804 atm. Find the value of the van't Hoff factor for magnesium chloride in this solution.

i = 2.483

Explanation:

The vapour pressure lowering formula is:

Pₐ = Xₐ×P⁰ₐ (1)

For electrolytes:

Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ

Where:

Pₐ is vapor pressure of solution (0.3624atm), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (0.3804atm)

4.5701g of MgCl₂ are:

4.5701g ₓ (1mol / 95.211g) = 0.048000 moles

43.238g of water are:

43.238g ₓ (1mol / 18.015g) = 2.400 moles

Replacing in (1):

0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm

0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)

2.4mol + i*0.048mol = 2.4mol / 0.9527

2.4mol + i*0.048mol = 2.5192mol

i*0.048mol = 2.5192mol - 2.4mol

i = 0.1192mol / 0.048mol

i = 2.483

I hope it helps!

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml