A 2.750×10−2M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving...

A 2.750×10⁻²M 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.

Part B

Calculate the mole fraction of salt in this solution.

Express the mole fraction to four significant figures.

Solutions

Expert Solution

mass of water = 999,3 x 0.9982

= 997.5 g

moles of water = 997.5 / 18 = 55.417 mol

moles of NaCl = 2.750×10⁻² x 1

= 2.750×10⁻²

molality = moles / mass of solvent in kg

= 2.750×10⁻² / 0.9975

molality = 0.0276 m

Part B)

mole fraction of salt = moles of salt / total moles

= 2.750×10⁻² / 2.750×10⁻² + 55.417

= 4.960 x 10^-4

queen_honey_blossom answered 1 year ago

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