An aqueous nacl solution is made using 112 g of nacl diluted to a total solution volume of 1.00 l. calculate the molarity, molality, and mass percent of the solution. (assume a density of 1.08 g>m

Q1)

molarity is defined as the number of moles of solute in 1 L of solution.

the NaCl solution volume is 1.00 L

number of moles NaCl = NaCl mass present / molar mass of NaCl

NaCl moles = 112 g / 58.5 g/mol = 1.91 mol

the number of moles of NaCl in 1.00 L of solution is - 1.91 mol

therefore molarity of NaCl is 1.91 M

Q2)

molality is defined as the number of moles of solute in 1 kg of solvent.

density is mass per volume.

density of the solution is 1.08 g/mL.

therefore mass of the solution is = density x volume

mass = 1.08 g/mL x 1000 mL = 1080 g

since we have to find the moles in 1 kg of solvent

mass of solvent = 1080 g - 112 g = 968 g

number of moles of NaCl in 968 g of solvent - 1.91 mol

therefore number of NaCl moles in 1000 g - (1.91 mol / 968 g) x 1000 g/kg = 1.97 mol/kg

molality of NaCl solution is 1.97 mol/kg

Q3)

mass percentage is the percentage of mass of solute by total mass of the solution

mass percentage of solution = mass of solute / total mass of the solution

mass of solute = 112 g

total mass of solution = 1080 g

mass % of NaCl = 112 g / 1080 g x 100%

therefore mass % of NaCl = 10.4 %

answer is 10.4 %