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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.

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Answer:

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!

The Van't Hoff factor is , i = 2.483. This can be calculated by using vapour pressure lowering formula.

The vapour pressure lowering formula is:

Pₐ = Xₐ*P⁰ₐ ...........(1)

For electrolytes:

Pₐ = nH₂O / (nH₂O + inMgCl₂)*P⁰ₐ

4.5701g of MgCl₂ are:

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

43.238g of water are:

43.238g * (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

Find more information about Van't Hoff factor here:

brainly.com/question/15134196

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