26.2k views
2 votes
A 1.30M solution of BaCl2 has a density of 1.230 g/ml. a) What is the mole fraction of BaCl2 in this solution?

1 Answer

4 votes

Answer: The mole fraction of barium chloride in the solution is 0.024

Step-by-step explanation:

We are given:

Molarity of barium chloride solution = 1.30 M

This means that 1.30 moles of barium chloride is present in 1 L or 1000 mL of solution.

  • To calculate the mass of solution, we use the equation:


\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.230 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:


1.230g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.230g/mL* 1000mL)=1230g

  • To calculate the number of moles, we use the equation:


\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} .....(1)

Moles of barium chloride = 1.30 moles

Molar mass of barium chloride = 208 g/mol

Putting values in equation 1, we get:


1.30mol=\frac{\text{Mass of barium chloride}}{208g/mol}\\\\\text{Mass of barium chloride}=(1.30mol* 208g/mol)=270.4g

Mass of water = Mass of solution - Mass of barium chloride

Mass of water = 1230 - 270.4 = 959.6 g

Calculating the moles of water:

Given mass of water = 959.6 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:


\text{Moles of water}=(959.6g)/(18g/mol)\\\\\text{Moles of water}=53.31mol

Mole fraction of a substance is given by:


\chi_A=(n_A)/(n_A+n_B)

  • For barium chloride:


\chi_{\text{(barium chloride)}}=\frac{n_{\text{(barium chloride)}}}{n_{\text{(water)}}+n_{\text{(barium chloride)}}}


\chi_{\text{(barium chloride)}}=(1.30)/(1.30+53.31)\\\\\chi_{\text{(barium chloride)}}=0.024

Hence, the mole fraction of barium chloride in the solution is 0.024

User Ziauz
by
7.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.