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21 July, 12:53

What are the molality and mole fraction of solute in a 22.3 percent by mass aqueous solution of formic acid (HCOOH) ?

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  1. 21 July, 12:59
    0
    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
  2. 21 July, 13:04
    0
    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
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