Answer:
Mole fraction for solute = 0.1, or 10%
Molality = 6.24 mol/kg
Explanation:
22.3% by mass → In 100 g of solution, we have 22.3 g of HCOOH
Mass of solution = 100 g
Mass of solute = 22.3 g
Mass of solvent = 100 g - 22.3g = 77.7 g
Let's convert the mass to moles
22.3 g . 1mol/ 46 g = 0.485 moles
77.7 g. 1mol / 18 g = 4.32 moles
Total moles = 4.32 moles + 0.485 moles = 4.805 moles
Xm for solute = 0.485 / 4.805 = 0.100 → 10%
Molality → mol/ kg → we convert the mass of solvent to kg
77.7 g. 1 kg / 1000g = 0.0777 kg
0.485 mol / 0.0777 kg = 6.24 m
Answer:
Molality = 6.23 molal
Mol fraction HCOOH = 0.101
Explanation:
Step 1: Data given
mass % of HCOOH = 22.3 %
Step 2: Calculate mass
22.3 % means we have 22.3 grams HCOOH in 100 gram solution
Mass of water = 100 grams - 22.3 grams = 77.7 grams
Step 3: Calculate moles
Moles HCOOH = mass HCOOH / molar mass HCOOH
Moles HCOOH = 22.3 grams / 46.03 g/mol
Moles HCOOH = 0.484 moles
Moles H2O = 77.7 grams / 18.02 g/mol
Moles H2O = 4.31 moles
Step 4: Calculate molality
Molality = moles HCOOH / mass H2O
Molality = 0.484 moles / 0.0777 kg
Molality = 6.23 molal
Step 5: Calculate mol fraction
Mol fraction HCOOH = mole HCOOH / total moles
Mol fraction HCOOH = 0.484 moles / (4.31+0.484)moles
Mol fraction HCOOH = 0.101
