Question

The density of a 15.00% by mass aqeous solution of acetic acid is 1.0181g/mL
What(i)the molarity (ii)molality (iii) the mole fraction of each component?

Answer 1

M = molarity = mols/L solution.
mass of 1 L of the solution is
mass = volume x density = 1000mL x 1.0181 = ?
mass of acetic acid (it's only 15% acid) = mass solution x 0.15 = ?
mols acetic acid = grams acetic acid/molar mass acetic acid = ?
That gives you the M since that is moles in 1 L solution.

m = molality = mols/kg solvent
You know mols acetic acid, you know grams of the solvent (that's grams 1 L solution - g acetic acid = grams water). Convert to kg water and m = mols acid/kg water.

X = mol fraction = mols acid/total mols
X = mols H2O/total mols.
You know mols acid from M work. You know mols H2O (mols = g/molar mass) from m work, substitute and solve for X of each. As a check, the Xwater + Xacid must be 1.000

Post your work. In the future, you need to show some work and/or tell us what you're thinking. This freebie stuff can't last long.