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

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

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

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.