1.What is the molarity of a sodium chloride solution, if 6.87 g of NaCl are diss

Question

1.What is the molarity of a sodium chloride solution, if 6.87 g of NaCl are dissolved in enough water to make 0.200 L of solution?

2. What is the molality of a sodium chloride solution, if 7.30 g of NaCl are dissolved in 0.900 kg of water?

3. What is the mole fraction of a lithium fluoride solution prepared by dissolving 11.2 g of LiF in 0.400 kg of water.

4. What is the mass percent of a lithium fluoride in a solution prepared by dissolving 50.7 g of LiF in 0.800 kg of water.

5. Calculate the freezing point for a 1.26 m solution of CCl4 in benzene.

What is the molarity of a sodium chloride solution, if 6.87 g of NaCl are dissolved in enough water to make 0.200 L of solution? What is the molality of a sodium chloride solution, if 7.30 g of NaCl are dissolved in 0.900 kg of water? What is the mole fraction of a lithium fluoride solution prepared by dissolving 11.2 g of LiF in 0.400 kg of water. What is the mass percent of a lithium fluoride in a solution prepared by dissolving 50.7 g of LiF in 0.800 kg of water. Calculate the freezing point for a 1.26 m solution of CCl4 in benzene.

Explanation / Answer

Moles of NaCl = 6.87/58.44=0.1176 moles

Molarity = 0.1176/0.200 =0.588M

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Moles of NaCl = 7.30/58.44 = 0.125moles

Molality = 0.125/0.900 = 0.139 molal

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Moles of LiF = 11.2/25.939=0.432

Moles of water = 0.4*1000/18 = 22.23

Mole fraction of LiF = 0.432/(0.432+22.23) = 0.019

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Mass percentage = 50.7/800 *100 = 6.3375%

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kf = 5.12

Depression in freezing point = 5.12*1.26 C = 6.45C

Freezing point of benzene = 5.5C

Freezing point of solution = 5.5-6.45 = -0.95 C

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