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

1 answer:

4 0

The correct answer is (C) 0.2 mole MgCl2.

according to Van't Hoff factor rule which is the ratio between the actual

concentration of the particles produced when the substance dissolved and

the concentration of the substance as measured from its mass, so here 0.2

mole MgCl2 is the higher van't Hoff factor as it has more ions than other

choices and the higher concentration.

and when the boiling point formula is delta Tb = Kb * m * i.

when delta Tb is the change in boiling point.

and Kb is molal boiling point constant

and m is the molality of the solution

and i is Van't Hoff factor.

so we can see the relationship between i and delta Tb, the greater Van't Hoff factor the higher the boiling point. so 0.2 MgCl2 is the higher boiling point as it is the greater van't Hoff factor.

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Ionization equation: CaCl2 ---> Ca (2+) + 2 Cl (-)

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formula of phospahte ion: PO4 (3-)

molar ratio: 2PO4(3-) + 3Ca(2+) = Ca3 (PO4)2

Proportion: 2 mol PO4(3-) / 3 mol Ca(2+) = x / 0.0825 mol Ca(2+)

=> x = 0.0825 coml Ca(2+) * 2 mol PO4(3-) / 3 mol Ca(2+) = 0.055 mol PO4(3-)

3) Content of Mg(2+) ions

Ionization equation: Mg (NO3)2 ----> Mg(2+) + 2 NO3 (-)

Molar ratios: 1 mol Mg(NO3)2 : 1 mol Mg(2+) + 2 mol NO3(-)

number of moles of Mg(NO3)2 in 1.5 liter of 8.0 * 10^-2 M solution

n = M * V = 8.0 * 10^ -2 M * 1.5 liter = 0.12 moles Mg(NO3)2

ions of Mg(2+) = 0.12 mol Mg(NO3)2 * 1 mol Mg(2+) / mol Mg(NO3)2 = 0.12 mol Mg(2+)

4) Number of phosphate ions needed to react with 0.12 mol Mg(2+)

2PO4(3-) + 3Mg(2+) = Mg3(PO4)2

=> 2 mol PO4(3-) / 3 mol Mg(2+) = x / 0.12 mol Mg(2+)

=> x = 0.12 * 2/3 mol PO4(3-) = 0.16 mol PO4(3-)

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=> 1 mol Na3PO4 : 1 mol PO4(3-)

=> 0.215 mol PO4(3-) : 0.215 mol Na3PO4

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=> mass in grams = 0.215 mol * 164 g/mol = 35.26 g

Answer: 35.26 g of sodium phosphate

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