[aficc]

Hey, I've been stuck on a problem that's from my chemistry textbook for quite a while so I've decided to come here to see if anyone can help. The answers are in the back of the book, (1. 0.634 L CO2, 2. k_0C = 7.61 x 10^-2 M/atm , k_20C = 3.84 x 10^-2 M/atm) so I'm just asking for how to approach/do it.  Here's the problem:

The solubility of CO_2 in water at 0°C and 1.00 atm is 0.335 g/100 g of H_2O. At 20.0° C and 1.00 atm, the solubility of CO_2 in water is 0.169 g/100 g of H_2O.

1. What volume of CO_2 would be released by warming 750 g of water saturated with CO_2 from 0° C to 20.0° C?

2. What is the value of the Henry's law constant for CO_2 under each set of conditions?

For #1, I tried using PV = nRT, where P = 1 atm, R = 0.0821, n = 0.169g x 7.5 / 44.1058, T = 293 K   and I got 0.69 L for volume... which is wrong.  For #2, Im pretty sure I have use the formula C = kP, (C = concentration, P = partial pressure, k = henry's constant.  However, I have no clue as to how to find C and P.

Any help is appreciated. Thank you.

[Dan]

1. You need to work out how much CO₂ is dissolved at the start of the experiment, and at the end. The difference is the amount of CO₂ released. Bear in mind that you will need to convert from grams of CO₂ into litres CO₂ at some point if you want to check your answer with the one in your book.

2.

p = pressure (given in the question)
k = constant (unknown)
c = concentration (you are given the mass of CO₂ dissolved at each temperature, and the mass of the solvent, so you can calculate c)

[aficc]

Sorry, but I'm still lost.  Ok so I calculate how much CO2 is dissolved at 0C, which is (7.5 x 0.335 = 2.5125g?) and at 20C, which is (7.5 x 0.169 = 1.2675g?). Is that right? If so, the difference is 1.245 g...? How do I convert it into litres?