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 %

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer