Question 45

[matthew]

I noticed that the osmotic pressure is basically the ideal gas law. In my mcat review book they have it rearranged. But really I need to know is the ideal and I can derive it during the CBT with g/m.w. for n and go from there.

For problem 45 just plugging in T got me 90,000 (some rounding). I almost thought thought T was delta T. Looking at my review book made sure the mistake doesn't happen on test day.

Doing all those thermo problems can get to you.

[sdsbama]

i'm having a mind blank here. How did you do the math so quickly and accurately in your head for doing 80/.00083?

[hulk475390]

80/.00083 = 80/(83 x 10-5) = 1 x 10^5 (some rounding)

[Guest]

its easier if you do fractions....

osmotic pressure = 2^-2
R(0.0x T (300) = 24 (memorize that)
So the math goes like this - no decimals, save long division for another day

2^-2=M24
2^-2/24 mol/L

You're given 80 g/L

80 g/L/1/12^-2 mol/L

multiply outside, put over inside
*remember your rules for division of exponents = ie switch the sign of the exponent and add it to the numerator

the fraction becomes

960^2/1 g/mol

[hua8986059]

I don't know how ppl finish these things on time, with all these calculations on 1 test..... I took AAMCs and there are like less than half the amount of calculations as this... lol.. and AAMC will often give you questions that just writes out equation as the answer for example if answer can be solved with Ideal gas law they will just give u PV = nRT as answer but with numbers plugged in and solving for one of the variables. U don't have to do the calculation part of it...

[dltkdgn896512]

easier way:

pie = MRT
0.0205 = M(.082)(300)

since .0205/.082 = 1/4

1/4 * (1/300) = M

1/1200 = M

1200 * 80 = 96000!!