can you teach me to calculate percent by mass? molarity? molality?

Question

pls pls pls teach me step by step....
i get confused whenever we have long test about this...

for example

water: 99 mL
H C2 H3 O2: 1 mL

THANKS IN ADVANCE....

Answer 1

Given what you've shown, you can't calculate any of that, because you don't know the mass or molarity of the 1 mL of HC2H3O2.

However, from Wikipedia, the density of liquid acetic acid is 1.049 g/cm^3 (or 1.049 g/mL)

Water's mass is 1 g/mL.

1.) % mass:
99 mL H20 * 1 g/mL = 99 g
1 mL C2H4O2 * 1.049 g/mL = 1.049 g

Total mass = 100.049
mass of C2H4O2 / total mass = 1.049 g / 100.049 g = 0.01048 = 1.048% C2H4O2 by mass. (solution!)

2.) Molarity = moles of solute / liter of solution
Step 1: Find molar mass of solute:
C2H4O2 = 2 C (12) + 4 H (1) + 2 O (16) = 60 g/mol

Step 2: Divide mass of solute by molar mass to find # of moles:
1 g / 60 g/mol = 0.0167 moles

Step 3: Divide # of moles by volume:
0.0167 moles / 100 mL = .167 moles/L = .167 M (solution!)

3.) Molality = # of moles / kg of solvent (not solution! In this case, the mass of the water):
# of moles = 0.0167 moles
mass of solvent = 99g = .099 kg
0.0167 moles / .099 kg = 0.169 molal