Osmotic Pressure Equation

[arc5005]

How many grams of a 12,000 gram/mole polymer must be added to water to make a 10-mL solution with an osmotic pressure of 0.0246 atm. at 27° C? (R = 0.082 L*atm/mole*K)

A. 0.012 grams polymer
B. 0.120 grams polymer
C. 1.200 grams polymer
D. 12.00 grams polymer

B) 0.120 grams polymer

I've seen the solution; however, I'm not sure I'd be able to exactly recognize and do this or a similar problem if it were to show up on an exam. Can anyone help me with this please?

---

Members don't see this ad.

 

[AdaptPrep]

Hi, arc5005-

The osmolarity of a solution is the molarity of the solution (M) multiplied by the number of ions the solute creates when dissolved in the solution (van't Hoff factor, I). Based on temperature of the solution, this osmolarity will produce a pressure across a membrane called osmotic pressure (pi). The formula is pi = iMRT where T is in Kelvin. Since this is a polymer, the value of i is one.
pi = iMRT
0.0246 = (1)(M)(0.082)(300)
0.0246 = 24.6M
0.001 = M

Since M (molarity) is moles per liter:
M = mole/L
0.001 = x/0.010
0.00001 moles = x
(12,000 grams/moles)(0.00001 moles) = 0.120 grams polymer

I hope that helps.

 

[arc5005]

Hi, arc5005-

The osmolarity of a solution is the molarity of the solution (M) multiplied by the number of ions the solute creates when dissolved in the solution (van't Hoff factor, I). Based on temperature of the solution, this osmolarity will produce a pressure across a membrane called osmotic pressure (pi). The formula is pi = iMRT where T is in Kelvin. Since this is a polymer, the value of i is one.
pi = iMRT
0.0246 = (1)(M)(0.082)(300)
0.0246 = 24.6M
0.001 = M

Since M (molarity) is moles per liter:
M = mole/L
0.001 = x/0.010
0.00001 moles = x
(12,000 grams/moles)(0.00001 moles) = 0.120 grams polymer

I hope that helps.

thank you so much!