What is the freezing point of an aqueous solution that boils at 103.7 degrees celsius?

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

What is the freezing point of an aqueous solution that boils at 103.7 degrees celsius?

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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

In this case, you need to take your equations for boiling point elevation and freezing point depression and combine them...\[\Delta T_b=K_b*m*i\]\[\Delta T_f=K_f*m*i\]Solve for (m * i) in the first equation, and then substitute that into the second to get...\[\Delta T_f=K_f*\frac{\Delta T_b}{K_b}\]K_f for water is 1.86 K/m, and K_b for water is 0.512K/m. \[\Delta T_f=(1.86 K*C^{-1})*\frac{103.7C-100C}{0.512K*C^{-1}}\]If we solve for deltaT_f, we get 13.4C, and since the normal freezing point for water is 0C, the depressed freezing point is...\[-13.4 C\]

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