Some Chem help, please...

<p>I have the ans. listed below. My question is how do you get to the answer?</p>

<h1>1 What is the mole fraction of KBr present in a solution that is 11.04% by mass aqueous KBr? The molar mass of KBr is 119.0 g/mol</h1>

<p>a. 0.0184 b. 0.05832 c. .1241 d. .2719</p>

<h1>2 Which of the following aqueous solutions has the lowest boiling point?</h1>

<p>a. 2M HCL b. 3 M ethly alcohol, C2H5OH c. 3 M HCL d. 2 M MgCl2</p>

<h1>3 Which solution has the highest pH?</h1>

<p>a. .1 M NH3 b. .1 M CH3COOH c. .1M NaOH d. .1M HCL e. .1 M Na2CO3</p>

<p>The ans.: 1. A 2. B 3 C.</p>

<p>Any idea on how to get started on these problems?
For #3, What I did was take the -log to get a pOH of 1. Then 14-1 to get a pH of 13. Is this the correct way of doing it?</p>

<p>Thanks in advance. Desperately need some assitance.</p>

<p>I know the answers to 2 and 3 and I'm being stupid and can't think of 1.</p>

<p>2) Use the boiling point elevation formula. You know molarity/molality and number of particles. Look up the Kbp (boiling point constant) and plug it into the formula.</p>

<p>3) You know that NaOH is the strongest base there and the stronger the base, the greater the pH value. Maybe there's another way to do it, but we haven't covered acids and bases yet this year.</p>

<p>Thanks. I appreciate it. For number 3, do you normally just focus on the NaOH and HCl since they are the strong acids and bases and disregard the rest?</p>

<p>Number 1:</p>

<p>(11.04/119.0)/(88.4/18.0)=0.0184</p>

<p>hey, s snack, just wondering, where did that 88.4 come from?
I know the 18 is 1 mol/18 g H2O.</p>

<p>100g-11.04g....(assume 100grams total solution)</p>

<p>You are correct for the third, even without further understanding the topic. If they gave you different molarities for the weak bases and NaOH you would actually have to look at the equillibrium systems and calculate the resulting pH. Obviously, a drop of a 10M weak base results in a more basic solution than a drop of a .1M strong base. However, in this case all the concentrations and amounts are the same, so all they are asking is for you to recognize that NaOH is the only strong base in the answer choices.</p>

<p>I see, thanks!</p>

<p>I have one more question:
Which of the following solutes dissolved in 1.0 kg of water would be expected to provide the fewest particles and FREEZE at the HIGHEST temperature (each solution contains 0.10 moles of solute)?
a. HClO3 b. HClO c. HCL d. HClO2 e. HClO4</p>

<p>Ans.: b. or d.</p>

<p>I'm pretty much stuck on this one. As for the three above, the help and input from you guys definately helped alot (and I know how to do them now :) )</p>

<p>I'm not sure of the answer since it would require looking up freezing point constants but you do know that D would dissolve more particles than B would (2 oxygen ions, 1 chloride, and 1 hydrogen for D as opposed to just 1 H+, 1 Cl-, and 1 O2-). The molality is mol solute/kg solvent so for all the molality is 0.1 mol/100 kg. Now look up the Kfp and plug it all into the equation. Whichever one gives you the smallest change in temperature (thus the highest freezing point) is the answer.</p>

<p>I'll do that then. This question was on the practice exam, and she did not give us the constants. So if this problem is really on there, she's suppose to give us the freezing point constants right?</p>

<p>Thanks for the help.</p>

<p>She should...unless she expects you to memorize the constants for every compound in existence...anyway, I hope this helps. Chemistry isn't by any means a strong subject for me...</p>

<p>Be careful with the aqueous ions for freezing point depression/ boiling point elevation problems. An ionic compound like HCl will ionize completely in water (into H+ and Cl-), but a molecular compound will not. CH3OOCH (acetic acid) + H2O --> CH3OOC(-)(aq) (the acetate ion) + H3O(+). The acetate ion does not ionize further.</p>

<p>You are right about not having to memorize constants, however. For the AP exam or SAT II you need to know the value of important constants of really common substances like Hydrochloric Acid, Sodium Hydroxide, and Acetic Acid. However, you will not be expected to know these values for every compound. They probably even give you the values for these common substances, but I would still recommend knowing them by heart.</p>

<p>I'll keep that in mind.</p>

<p>I have a few more that I'm totally stuck.
1. A solution that is formed by combining 200.0 mL of 0.15 M HCl with 300.0 mL of 0.090M NaOH has an OH- concentration of
a. 0.0030 b. 3.3 x 10^-12 c. 1.7 X10^-12 d. 0.030 e. 1.0 x 10^-7</p>

<ol>
<li><p>Which of the following liquids is most likely to be miscible with water?
a. CH2Cl2 b. C6H14 c. CCl4 d. acetic acid </p></li>
<li><p>What chemical process would liberate the most energy
a. O (g) + e^- -->O^-(g)
b. O^-1(g) + e^-1 --> O^-2 (g)
c. O^-2 (g) + e^- --> O^-3(g)<br>
d. O(g)--> O^+(g) +e^-</p></li>
</ol>

<p>Ans: 1. A 2. D 3. A</p>

<p>For #2, is it acetic acid (D) because it's a gas?</p>

<p>Sorry for bugging you guys, but chem is my weakest of the three sciences.</p>

<ol>
<li><p>Acetic Acid = CH3COOH. It's polar (it has carboxyl group -COOH), and like will dissolve like (water is also polar).</p></li>
<li><p>A, because the first electron affinity for msot nonmetal (adding an electron to a mole of the atom) will release energy. D is just wrong. And B and C require energy. Why? Because it takes energy to put an electron in a negatively charge ion (it's not gonna want to go).</p></li>
</ol>