Question

Chemists often use molarity M, in moles/liter, to measure the concentration of solutions. Molarity is a common unit of concentration because the volume of a liquid is very easy to measure. However, the drawback of using molarity is that volume is a temperature-dependent quantity. As temperature changes, density changes, which affects volume. Volume markings for most laboratory glassware are calibrated for room temperature, about 20∘C.

A 2.500×10⁻²M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.3 mL . The density of water at 20.0∘C is 0.9982 g/mL.

Part A

Calculate the molality of the salt solution.

Answer: mNaCl = 2.506×10⁻² m

Part B

Calculate the mole fraction of salt in this solution.

Express the mole fraction to four significant figures.

Answer: ?????

Part C

Calculate the concentration of the salt solution in percent by mass.

Express your answer to four significant figures and include the appropriate units.

Answer:?????

Part D

Calculate the concentration of the salt solution in parts per million.

Express your answer as an integer to four significant figures and include the appropriate units.

Answer:??????

*Just need help with part b,c, and d. Please and thank you!*

Answer 1

moles of NaCL present= 2.500×10−2 mols

mass of water present = volume*density
=999.3 mL *0.9982 g/mL = 997.5 g
moles of water present = 997.5 g/18.015 g/mol= 55.37 mol

mol fraction of NaCl = moles of NaCL present/total number of mols
= 2.500×10−2 mols/2.500×10−2 mols+55.37 mol
=2.500×10−2 /55.395
= 4.51 x10^-4
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Part C

mass of solute NaCL = moles of NaCl*molar mass of NaCl
= 2.5 x10^-2 moles*58.44 g/mol = 1.461 g
mass of water = 997.5 g

mass % = mass of solute/total mass*100 = [1.461/1.461+997.5] *100
=0.146 %
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1ppm = 1 mg/L
mass of solute NaCl = 1.461 g = 1461 mg

So concentration in ppm = 1461 ppm
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