Question

# A 2.650×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by...

A 2.650×10−2 M 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.

Calculate the mole fraction of salt in this solution.

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

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

First we have to calculate

moles of NaCl = 2.65 x 10^-2 mol

mass of water = 999.3 ml x 0.9982 g/mol/1000

Water mass = 0.9975 kg

1.

To determine the mole fraction of salt

Consider,

moles water = 997.5 g/18 g/mol

moles water = 55.42 mol

complete moles = 55.42 + 0.0265

= 55.45 mol

mole portion of salt in arrangement = 0.0265/55.45

mole fraction of salt in the solution= 0.0005

2.

To determine concentration of salt solution in percent by mass

mass NaCl in arrangement = 2.65 x 10^-2 mol x 58.44 g/mol

= 1.55 g

Complete mass of arrangement = 1.55 + 997.5 g

= 999.05 g

mass% NaCl in arrangement = 1.55 x 100/999.05

concentration of the salt solution in percent by mass = 0.155%

3.

To determine salt solution concentration in ppm

NaCl focus in ppm = 1.55 g x 1000/1 L

concentration of the salt solution in parts per million = 1550 ppm.

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