Respuesta :
Answer: The mole fraction of barium chloride in the solution is 0.024
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:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of solution = 1.230 g/mL
Volume of solution = 1000 mL
Putting values in above equation, we get:
[tex]1.230g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.230g/mL\times 1000mL)=1230g[/tex]
- To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Moles of barium chloride = 1.30 moles
Molar mass of barium chloride = 208 g/mol
Putting values in equation 1, we get:
[tex]1.30mol=\frac{\text{Mass of barium chloride}}{208g/mol}\\\\\text{Mass of barium chloride}=(1.30mol\times 208g/mol)=270.4g[/tex]
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:
[tex]\text{Moles of water}=\frac{959.6g}{18g/mol}\\\\\text{Moles of water}=53.31mol[/tex]
Mole fraction of a substance is given by:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex]
- For barium chloride:
[tex]\chi_{\text{(barium chloride)}}=\frac{n_{\text{(barium chloride)}}}{n_{\text{(water)}}+n_{\text{(barium chloride)}}}[/tex]
[tex]\chi_{\text{(barium chloride)}}=\frac{1.30}{1.30+53.31}\\\\\chi_{\text{(barium chloride)}}=0.024[/tex]
Hence, the mole fraction of barium chloride in the solution is 0.024
