Sat II Chem question

Test Preparation SAT Subject Test Preparation
1

<p>Question 42 in barrons test 3
When 1 mole of NaCl is dissolved in 1000 g of water, the boiling point of the water is change to...
a. 100.51 C
b. 101.02 C
c. 101.53 C
d. 101.86 C
e. 103.62 C
okay, so you'd used the KMI equation.. but are you just supposed to know that k is .51? and is it always .51? or only for NaCl...
The answer is B.</p>

<p>also;
69. what is the molar mass od a nonionizing solid if 10 g of this solid, dissolved in 200 grams of water, formed a solution that froze at -3.72 C?
a. 25 g/mol
b. 50
c. 100
d.150
e.1000</p>

<p>the answer is A.</p>

2

<p>Yeah i noticed that also, i think it's just plain ridiculous that they expect you to know the constant. by the way, that's the only way you can get the answer: know the constant. CB don't even expect you to know on the AP exam, why would they expect you to know this on the SATII, which involves much less calculation?</p>

3

<p>so do we have to know the specifics? such as that constant. and do we have to know the name of each chemist that made a law or theory?</p>

4

<p>@Msu</p>

<p>I keep getting 6.85 g/mol for #69. Also, they wouldn't give you this particular problem (69) because of all the math involved without a a calculator.</p>

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<p>what kind of problem is that though?</p>

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<p>If you knew the constant, the math in #69 is pretty simple.</p>

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<p>@I<3Pi</p>

<p>3.72=X moles * (0.5)/0.2
Dividing 3.72 by 5 isn't too hard but still... dividing by 0.5 (multiply by 2) is easy. You get 1.488 moles</p>

<p>Now, set y(molecular weight)=10 g/1.488 moles</p>

<p>Umm...you do this by hand and come back to me. Keep in mind the time limits.</p>

8

<p>Hey afruff. well the molar freesing constant is -1.86 something(I think cal) per mole. So I think through inspection one could see that -3.72 is twice -1.86. Which means there are two moles of solute in solution. Again through simple proportions we see that there are 50 g in 1000mL. Now remember this is a 2 molar soultion so 50/2 = 25...A</p>

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<p>Oh, I used 0.5 as K since that's what MSU said.</p>

10

<p>thats because one is boiling point and ones freezing point</p>

11

<p>Oh, my bad. I should pay more attention.</p>

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<p>@afruff23 </p>

<p>I did do it. I did test #3 in Barron's yesterday. What I did: since the constant is -1.86, you know that the solution must be 2 molal (-3.72 / 2). You have 0.2kg of the solution, which means 0.4 moles in the solution. You have 10g, 10g/0.4mol = 25g/mol. See? Easy.</p>

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<p>I was using 0.5 which does not divide into 3.72 so well. If you use -1.86 then it is definitely doable.</p>