Ten (10.0) grams of sodium chloride, NaCl, are added to 1.00 L of water. Ten (10

Question

Ten (10.0) grams of sodium chloride, NaCl, are added to 1.00 L of water. Ten (10.0) grams of glycerol, C3H8O3, are added to 1.00 L of water.

None of the above statements are true.

Given the following information: A sulfuric acid solution contains 5.80 x 102g of H2SO4per liter of solution. The density of the solution is 1.329 g/mL. The mole fraction of the water in the solution is

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Question 13 10 pts

Given the following information: A sulfuric acid solution contains 5.80 x 102 g of H2SO4 per liter of solution. The density of the solution is 1.329 g/mL. The molality of the solution is

The freezing point of the sodium chloride solution will be identical to the freezing point of the glycerol solution.

Explanation / Answer

10 g NaCl / 58.44 = 0.171 mol NaCl X 2 = 0.342 mol ions in solution

10 g glycerol / 92.01 = 0.109 mol glycerol

So, The freezing point of the sodium chloride solution will be lower than the freezing point of the glycerol solution.

Question 2

Molar mass of H2SO4 = 98.1 g/mol

H2SO4 Moles = 5.80 X 102 g X 98.1 g/mol = 5.91 mol H2SO4

The mass of water in this 1 L of solution = 1329 - 580 = 749 g = 0.749 kg

Moles of water = 749 / 18 = 41.61 Moles

Mole fraction of water = 41.61 / 41.61 + 5.91 = 0.8756

Question 3

Molar mass of H2SO4 = 98.1 g/mol

H2SO4 Moles = 5.80 X 102 g X 98.1 g/mol = 5.91 mol H2SO4
The mass of this solution 1 L of solution is: 1000 mL X 1.329 g/mL = 1329 g
The mass of water in this 1 L of solution = 1329 - 580 = 749 g = 0.749 kg

So, the molality of the solution is:
5.91 mol solute / 0.749 kg = 7.9 molal

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