What are the molality and mole fraction of solute in a 30.1 percent by mass aqueous solution of formic acid (HCOOH)
In a 100 g sample, you will have 30.1 g HCOOH and 69.9 g H2O.
mols HCOOH = 30.1/molar mass = ?
Then m = mols HCOOH/0.1 kg solvent = ?
XHCOOH = mols HCOOH/total mols.
You have mols HCOOH. mols H2O \ grams H2O/molar mass = ?
To find the molality and mole fraction of the solute in a solution, we need to first determine the grams of solute and solvent present.
Given:
- Mass percent of formic acid (HCOOH) in the solution: 30.1%
- We'll assume we have 100 grams of the solution.
To calculate the grams of formic acid in the solution, we can use the mass percent:
Mass of formic acid = (mass percent / 100) * mass of solution
= (30.1 / 100) * 100 grams
= 30.1 grams
Since we know the mass of the solution is 100 grams, the mass of water (solvent) can be calculated as:
Mass of water = mass of solution - mass of formic acid
= 100 grams - 30.1 grams
= 69.9 grams
Next, we'll calculate the molality of the solute:
Molality (m) = (moles of solute) / (mass of solvent in kg)
To find the moles of solute, we need to convert grams of formic acid to moles using its molar mass. The molar mass of formic acid (HCOOH) is:
Molar mass (HCOOH) = (1*1.008) + (1*12.01) + (1*16.00) + (1*1.008)
= 46.03 g/mol
Moles of formic acid (HCOOH) = mass of formic acid / molar mass (HCOOH)
= 30.1 grams / 46.03 g/mol
= 0.653 mol
Now, we need to convert the mass of water to kilograms:
Mass of water (kg) = mass of water / 1000
= 69.9 grams / 1000
= 0.0699 kg
Finally, we can substitute the values into the molality equation:
Molality (m) = 0.653 mol / 0.0699 kg
= 9.35 mol/kg
So, the molality of the solute (formic acid) in this solution is 9.35 mol/kg.
Lastly, let's calculate the mole fraction of the solute:
Mole fraction (X) = (moles of solute) / (moles of solute + moles of solvent)
Moles of solvent (water) = mass of water / molar mass (H2O)
= 69.9 grams / 18.01 g/mol
= 3.88 mol
Mole fraction (X) = 0.653 mol / (0.653 mol + 3.88 mol)
= 0.144
Therefore, the mole fraction of the solute in the solution is 0.144.