Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Calculate the mole fraction of HCL, MM = 36.46 g/mol , in a 1.20 M aqueous, MM H20 = 18.02 g/mol , solution when a density is 1.18 g/mol 25 degree C.

Chemistry
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

You know that 1L of the solution contains 1.20 moles HCl, to find the mole fraction, you must find how many moles of water the solution contains, and then divide the number of moles HCl by the total number of moles present in the solution. Here you have that 1 L of solution contains 1.20 mol HCl and it weighs 1.18g*1000mL=1180 g. The weight of water is 1180g-(36,46g/mol*1.20 mol HCl)=1180 g-43.75g=1136.25g water. Then divide the weight of the water by the Mm of H2O to find the number of moles H2O: 1136/18.02=63.05 mol. The last step is to divide the number of moles HCl by the total number of moles present in the solution: 1.20/(1.20+63.05)=0.1867, which is approximately 0.0187.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question