Question

What is the mass percent of the solution. if An aqueous NaCl solution is made using 138 g of NaCl diluted to a total solution volume of 1.10 L Assume a density of 1.08 g per ml for the solution.?

Answer 1

MV = mass / molar mass 

(x) (1.30 L) = 138 g / 58.443 g/mol 

x = 1.82 M 

Part B: 

molality is moles solute per kg of solvent. I will use 2.3613 mol (keeping a few guard digits). 

Let us assume 1000 mL of the solution is present. This tells us that 2.3613 mol of the solute is present (that's the 138 g of NaCl). 

1.08 g/mL times 1000 mL = 1080 g <--- this is the total mass of the 1000 mL solution 

1080 g minus 138 g = 942 g <--- the mass of water in the 1000 mL of solution 

942 g = 0.942 kg 

2.3613 mol / 0.942 kg = 2.51 m 

Part C: 

138 g of solute was dissolved in 1080 total grams of solution 

(138 / 1080) * 100 = 12.8% <--- NaCl 

100% minus 12.8% = 87.2% <--- H2O 

Edit: another type of question is this area is to ask you to determine the mole fractions of each substance. For that you will need to know the moles of water: 

942 g / 18.015 g/mol = 52.29 mol 

The mole fraction of NaCl is this: 

2.3613 mol / (2.3613 mol + 52.29 mol) = 0.0432 

The mole fraction of the water is this: 

1 - 0.0432 = 0.9568