Question

What are the molality and mole fraction of solute in a 35.9 percent by mass aqueous solution of formic acid (HCOOH)?

Answer 1

ِAnswer:

1- The molarity of HCOOH = 9.515 M.

2- The mole fraction of HCOOH = 0.18.

Explanation:

1- The molarity of HCOOH:

• We can calculate the molarity of HCOOH using the relation:

M = (10pd)/molar mass.

p is the percent by mass of HCOOH = 35.9 %.

d is the specific gravity of HCOOH = 1.22 g/cm³.

Molar mass of HCOOH = 46.03 g/mol.

∴ M = (10pd)/molar mass = (10)(35.9 %)(1.22 gcm³) / (46.03 g/mol) = 9.515 M.

2- The mole fraction of HCOOH:

• We can suppose that we have a 100 g solution, that contains 35.9 g of HCOOH and 64.1 g of water.

The mole fraction of HCOOH = (no. of moles of HCOOH) / (no. of moles of HCOOH + no, of moles of water).

no. of moles of HCOOH = mass / molar mass = (35.9 g)/(46.03 g/mol) = 0.78 mol.

no. of moles of water = mass / molar mass = (64.1 g)/(18.0 g/mol) = 3.56 mol.

• The mole fraction of HCOOH = (no. of moles of HCOOH) / (no. of moles of HCOOH + no, of moles of water) = (0.78 mol) / (0.78 mol + 3.56 mol) = 0.18.