Question 6 Review

Which sample, when dissolved in 1.0 liter of water, produces a solution with the highest boiling point?

A

0.1 mole Ki

B

0.2 mole KI

0.1 mole MgCl2

0.2 mole MgCl2

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0.2 mol MgCl₂

Explanation:

ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of pure solvent

Kb is a constant. For water is 0.52 °C/m

m = molality (moles of solute in 1kg of solvent)

i = numbers of ions dissolved.

That's all the elements for the formula to determine the elevation of boiling point.

We assume, that molality will be at least the same for salts with 0.1 moles and 0.2 moles. For 0.1 mol :

1L of water contains 1000g of it, so we can also say that 1kg of water is contained in 1 L of water. Remember that density is 1g/mL or 1kg/L

So concentrations for 0.1 moles are, 0.1 mol/kg and for 0.2moles, 0.2 mol/kg. Let's dissociate the salts to determine i:

KI  → K⁺  +  Cl⁻     i = 2

MgCl₂  → Mg²⁺  +  2Cl⁻    i = 3

The highest difference will be at the 0.2 moles of MgCl₂, so that will be the sample that produces a solution with the highest boiling point.

Boiling T° of solution = 0.52 °C/m . 0.1m . 2  + 100°C = 100.10°C

Boiling T° of solution = 0.52 °C/m . 0.2m . 2  + 100°C = 100.20°C

Boiling T° of solution = 0.52 °C/m . 0.1m . 3  + 100°C = 100.15°C

Boiling T° of solution = 0.52 °C/m . 0.2m . 3  + 100°C = 100.31°C

answer: mole of the gas

answer: mole of the gasexplanation: