What are the molality and mole fraction of solute in a 25.5 percent by mass aqueous solution of formic acid (HCOOH)?
To find the molality and mole fraction of a solute in a solution, you will need the masses and molecular weights of the solute and solvent. In this case, the solute is formic acid (HCOOH) and the solvent is water (H2O).
Step 1: Determine the masses of solute and solvent
In a 25.5% by mass solution, 25.5 g of the solution contains 100 g of the solution. Let's assume we have 100 g of the solution. Therefore, the mass of the solute (formic acid) is 0.255 * 100 g = 25.5 g. The mass of the solvent (water) is 100 g - 25.5 g = 74.5 g.
Step 2: Calculate the number of moles of solute
To determine the number of moles of solute, we need to know the molecular weight of formic acid (HCOOH). The molecular weight of HCOOH is approximately 46.03 g/mol.
Number of moles of solute = Mass of solute / Molecular weight of solute
= 25.5 g / 46.03 g/mol
≈ 0.554 mol
Step 3: Calculate the molality of the solute
Molality is defined as the number of moles of solute per kilogram of solvent. Since we have calculated the number of moles of solute in the previous step, we need to convert the mass of the solvent from grams to kilograms.
Mass of solvent (H2O) = 74.5 g = 0.0745 kg
Molality of solute = Number of moles of solute / Mass of solvent (in kg)
= 0.554 mol / 0.0745 kg
≈ 7.44 mol/kg
Step 4: Calculate the mole fraction of solute
The mole fraction of solute is the ratio of moles of solute to the total moles in the solution. To find the total moles, we need to calculate the number of moles of water.
Number of moles of water = Mass of water / Molecular weight of water
≈ 74.5 g / 18.02 g/mol
≈ 4.13 mol
Mole fraction of solute = Moles of solute / (Moles of solute + Moles of water)
= 0.554 mol / (0.554 mol + 4.13 mol)
≈ 0.118
Therefore, the molality of the solute in the 25.5% by mass aqueous solution of formic acid is approximately 7.44 mol/kg, and the mole fraction of solute is approximately 0.118.
To find the molality and mole fraction of solute in a solution, we need to know the mass, molar mass, and density of the solution.
To find the molality of the solute, we first need to determine the mass of the solute in the solution. Given that the solution is 25.5% by mass, we know that 25.5 grams of formic acid (HCOOH) are dissolved in every 100 grams of solution. So, if we consider 100 grams of solution, we will have 25.5 grams of formic acid.
Next, we need to calculate the number of moles of formic acid using its molar mass. The molar mass of formic acid is 46.03 g/mol. We can use this value to determine the number of moles by dividing the mass of formic acid by its molar mass:
Number of moles = Mass of formic acid / Molar mass
= 25.5 g / 46.03 g/mol
≈ 0.554 mol
The molality (m) of the solution is defined as the number of moles of solute per kilogram of solvent. Since the solvent is water, we need to convert the mass of solvent (100 g) to kilograms (0.1 kg):
Molality = Number of moles of solute / Mass of solvent (in kg)
= 0.554 mol / 0.1 kg
= 5.54 mol/kg
So, the molality of the solution is 5.54 mol/kg.
To find the mole fraction of the solute, we need to consider the total number of moles in the solution. The total number of moles is the sum of the moles of solute and moles of solvent (water in this case).
Since the mass percent composition is given, we know that there are 25.5 grams of formic acid (HCOOH) in every 100 grams of solution. The remaining mass (100 g - 25.5 g = 74.5 g) is the mass of water.
To find the number of moles of water, we divide the mass of water by its molar mass:
Number of moles of water = Mass of water / Molar mass of water
= 74.5 g / 18.02 g/mol
≈ 4.13 mol
The total number of moles in the solution is the sum of the moles of water and the moles of formic acid:
Total number of moles = moles of water + moles of formic acid
= 4.13 mol + 0.554 mol
≈ 4.68 mol
The mole fraction (χ) of the solute (formic acid) is given by:
Mole fraction = Moles of solute / Total number of moles
= 0.554 mol / 4.68 mol
≈ 0.118
So, the mole fraction of the solute in the solution is approximately 0.118.