The density of a 15.00% by mass aqueous solution of acetic acid, CH3COOH, is 1.0187g/mL. What is (a) the molarity? (b) the molality? (c) the mole fraction of each component?

density = 1.0187 g/mL.

mass 1000 mL = 1.0187 x 1000 mL = ??
How much of that is acetic acid? 15%; therefore, ?? x 0.15 = mass acetic acid = xx.
How many mols is that?
xx g acetic acid/molar mass acetic acid = yy
Molarity = yymols/L = yy M.

molality = mols solute/kg solvent
mols you have from above. kg solvent is mass of 1 L solution which is 1018.7 g- mass acetic acid = mass H2O remaining.

mol fraction.
mols acetic acid = from above.
mols H2O = g H2O/molar mass H2O.

mol fraction acetic acid = mols acetic acid/sum mols acetic acid + mols H2O.

mols fraction H2O = etc.

Check my work. Check my thinking.

thanks so much for your help!!

i think i understood everything you said...but just in case am i on the right track with these answers?

molarity = 2.545M
molality = 2.939m
mole fraction: acetic acid = .05030
H20 = .9497

I remember working the molarity and molality and those are correct if my memory isn't failing me. I just worked the mole fraction one, but rounded on molar masses etc, and your numbers agree at least to three significant figures. So I think you have it ok. Glad to help. Come back anytime.

a 7.50 by mass aqueous ammonia, NH3,solution has a density of 0.973 g/mol. Calculate molality and molarity? Need your ASAP please :(

To answer these questions, we need to understand the definitions of molarity, molality, and mole fraction.

(a) Molarity (M) is defined as the number of moles of solute per liter of solution. It is expressed in moles per liter (mol/L).

(b) Molality (m) is defined as the number of moles of solute per kilogram of solvent. It is expressed in moles per kilogram (mol/kg).

(c) Mole fraction (χ) is defined as the ratio of moles of a component to the total moles of all components in the solution. It is a dimensionless quantity that ranges from 0 to 1.

To calculate each of these quantities, we need to know the molar mass of acetic acid, CH3COOH. The molar mass of carbon (C) is 12.01 g/mol, hydrogen (H) is 1.01 g/mol, and oxygen (O) is 16.00 g/mol.

Step 1: Calculate the molar mass of acetic acid (CH3COOH).
Molar mass = (2 × C) + (4 × H) + (2 × O)
Molar mass = (2 × 12.01) + (4 × 1.01) + (2 × 16.00)
Molar mass = 60.05 g/mol

Step 2: Calculate the mass of acetic acid in the solution.
Given that the solution is 15.00% by mass, we can assume we have 100 g of the solution. This means there are 15.00 g of acetic acid in the solution.

Step 3: Calculate the volume of the solution.
Given that the density of the solution is 1.0187 g/mL, we can assume we have 100 mL of the solution. This means we have 100 g of the solution, as 1 mL of water has a mass of approximately 1 g.

(a) The molarity (M) can be calculated as follows:
Molarity (M) = moles of solute / volume of solution in liters
moles of solute = mass of acetic acid / molar mass of acetic acid
= 15.00 g / 60.05 g/mol
= 0.2498 mol

volume of solution = mass of solution / density of solution
= 100 g / 1.0187 g/mL
= 98.18 mL = 0.09818 L

Molarity (M) = 0.2498 mol / 0.0982 L
= 2.54 M

Therefore, the molarity of the solution is 2.54 M.

(b) To calculate the molality (m), we need to know the mass of the solvent (water) in kilograms.
mass of solvent = mass of solution - mass of solute
= 100 g - 15.00 g
= 85.00 g = 0.08500 kg

Molality (m) = moles of solute / mass of solvent in kg
= 0.2498 mol / 0.0850 kg
= 2.94 mol/kg

Therefore, the molality of the solution is 2.94 mol/kg.

(c) The mole fraction (χ) of acetic acid can be calculated as follows:
moles of acetic acid = mass of acetic acid / molar mass of acetic acid
= 15.00 g / 60.05 g/mol
= 0.2498 mol

moles of water = mass of solvent / molar mass of water
= 85.00 g / 18.02 g/mol
= 4.71 mol

Total moles = moles of acetic acid + moles of water
= 0.2498 mol + 4.71 mol
= 4.96 mol

Mole fraction of acetic acid (χ acetic acid) = moles of acetic acid / total moles
= 0.2498 mol / 4.96 mol
= 0.0503

Mole fraction of water (χ water) = moles of water / total moles
= 4.71 mol / 4.96 mol
= 0.9497

Therefore, the mole fraction of acetic acid is approximately 0.0503, and the mole fraction of water is approximately 0.9497.