What are the molality and mole fraction of solute in a 30.1 percent by mass aqueous solution of formic acid (HCOOH)

can you please explain it better? I still don't quite understand

assume 1000g of solution

so 301 grams of it is formic acid, and 699 grams of water.

moles of water=699/18= ...
moles of formic acid=301/molmass=..

mole fraction=molesacid/totalmoles
molality= molesacid/.699

Daryll, "Explain it further" doesn't help much. Did you try following each of the formulas I gave. Exactly what do you not understand. Do it step by step, show your work, and explain each part you don't understand.

Sure! Let's break it down step by step:

1. Start by converting the given percentage by mass (30.1%) into grams. This means that in 100 grams of the solution, 30.1 grams is made up of formic acid.

2. Next, calculate the molar mass of formic acid (HCOOH). It is 46.02 grams/mol.

3. Now, divide the mass of formic acid (30.1 grams) by the molar mass (46.02 grams/mol) to find the number of moles of formic acid in the solution.

Moles of formic acid = 30.1 g / 46.02 g/mol

4. Now, determine the mass of water in the solution. Since the solution is aqueous, we assume water is the solvent. Therefore, the remaining mass of the solution (100 g - 30.1 g) is the mass of water, which is 69.9 grams.

5. Calculate the molality (m) of the solution by dividing the moles of formic acid by the mass of water in kilograms.

Molality (m) = Moles of formic acid / Mass of water (in kg)

6. Determine the mole fraction (χ) of the solute (formic acid) in the solution. The mole fraction is the ratio of the moles of the solute to the total moles of all components in the solution.

Moles of solute = Moles of formic acid
Moles of solvent = Moles of water

Mole fraction (χ) of solute = Moles of solute / (Moles of solute + Moles of solvent)

These calculations will give you the molality and mole fraction of the solute (formic acid) in the given solution.