Question

A
0.02350 M solution of NaCl in water is at 20degrees. 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 999.4ml The density of water @20
degrees is 0.9982g/mL.

Calculate the morality of the salt solution, the mole fraction
of salt in this solution, the concentration of the salt solution in
% by mass, & the concentration of the salt solution in parts
per million?

Answer #1

1,Mass of water in 999.4ml = 0.9982*999.4 gms=997.6 gms

Moles of water = 997.6/18=55.42

Moles of NaCl in 0.02350 in 1 L =0.02350 moles

Mass of NaCl =0.02350*58.5=1.374 gms

Molarity = moles of solute/ Liter of solution =0.02350M

2. Total moles = moles of water + moles of NaCl =55.42+0.02350=55.4435

Mole fraction= moles of NaCl/ Moles of solution= 0.02350/55.4435=0.000424

3. Total mass = mass of water + mass of NaCl = 997.6+1.375

=998.975 gms

Mass % of NaCl = 1.375/998.74=0.001376

4. mass of NaCl =1.375 gm/ 998.975gm
=0.001376*10^{6}/10^{6} =1376 ppm

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