Ethylene glycol (C2H6O2) is used as an antifreeze in cars. If 250 g of ethylene glycol is added to 3.00 kg of water, what is the molality? Calculate how much the freezing point of water will be lowered. The freezing-point depression constant for water is Kf = –1.86°C/m. Show your work.

2,909 M

Explanation:

molair mass is of.ethylene is 26,04 g/mol

first you need to calculate how much mL 3 kg is. You can do this by using the density of ethylene: 1,1 g/mL.

3000 g x 1.1 = 3300 mL = 3,3 L

Next you need to calculate the amount of moles:

250 g / 26,04 g/mol = 9,60 mol

Now you can calculate the molarity:

9,6/3.3 = 2,909 M

I don't know the answer for the second question. I'm sorry.

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.