For an aqueous solution of sodium chloride (nacl), determine the molarity of 4.15 l of a solution that contains 173 g of sodium chloride. determine the volume of this solution that would contain 3.93 moles of sodium chloride. determine the number of moles of sodium chloride in 22.45 l of this solution.

a) Molarity = 0.713 M

b) volume = 5.51 L

c) Number of moles = 16.01 moles

Explanation:

A) What is the molarity of a 4.15L solution that contains 173 g of sodium chloride?

Step 1: Data given

Volume = 4.15 L

Mass of NaCl = 173 grams

Molar mass of NaCl = 58.44 g/mol

Step 2: Calculate moles of NaCl

Number of moles NaCl = Mass NaCl / molar mass

Moles NaCl = 173 grams / 58.44 g/mol

Moles NaCl = 2.96 moles

Step 3: Calculate molarity of solution

Molarity = moles NacL/ volume

Molarity = 2.96 moles / 4.15L

Molarity = 0.713 M

b) Determine the volume of this solution that would contain 3.93 moles of sodium chloride.

Step 1: Data given

Number of moles = 3.93 moles

Molarity = 0.713 M

Step 2: Calculate volume

Volume = Moles/ Molarity

Volume = 3.93 mol/0.713 M

volume = 5.51 L

c) Determine the number of moles of sodium chloride in 22.45 L of this solution.

Step 1: Data given

Volume = 22.45 L

Molarity = 0.713 M

Step 2: Calculate number of moles

Moles NaCl = Molarity * volume

Moles NaCl = 0.713 * 22.45 L

Moles NaCl = 16.01 moles

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