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

Question

College

Thank you for visiting What are the molality and mole fraction of solute in a 35 9 percent by mass aqueous solution of formic acid HCOOH. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!

Answer :

superman1987 superman1987 · Answer 1 · 2024-12-16T21:49:50-05:00

ÙAnswer:

1- The molarity of HCOOH = 9.515 M.

2- The mole fraction of HCOOH = 0.18.

Explanation:

1- The molarity of HCOOH:

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.