anonymous
  • anonymous
1. Determine the osmotic pressure of a solution that contains 0.020 g of a hydrocarbon solute (molar mass = 340 g/mol) dissolved in benzene to make a 350-mL solution. The temperature is 20.0°C. A)3.1 torr B)1.0 torr C)2.9 torr D)0.21 torr E)1.4 torr 2. Which of the following solutions would have the highest osmotic pressure? A)0.2 M C6H12O6, glucose B)0.15 M MgBr2, magnesium bromide C)0.2 M CH3OH, methanol D)0.2 M C12H22O11, sucrose E)0.15 M KBr, potassium bromide
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
1. PV=nRT 2. Osmotic pressure=molarity multiplied by the number of ionic element. Ex: NaCl has 2 \[Na ^{+}, Cl ^{-}\] \[AlCl _{3} \] has 4 ions \[Al ^{3+},Cl ^{-},Cl ^{-}, Cl ^{-}\]
anonymous
  • anonymous
The equation is not working for me, but do you understand?
anonymous
  • anonymous
P is pressure in atm V is volume in liters n is the number of moles R is a constant which is 0.082 Temperature is in Kelvin.

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aaronq
  • aaronq
The formula for osmotic pressure is: \(\huge \sf\Pi=iMRT \) \(\sf \Pi\) is Osmotic pressure \(\sf i\) is the van't hoff constant \(\sf M\) is molarity \(\sf R\) is the gas constant \(\sf T\) is the absolute temperature
anonymous
  • anonymous
Sorry, I meant to wrote osmotic pressure formula for this problem. We do not need to know the temperature or r because they are the same.
aaronq
  • aaronq
For 2. you dont need it - clearly it's dependent on \(i\), but for 1. you do need to use the whole formula including the temperature.
anonymous
  • anonymous
Yes, I referred to 2 only since that is the one asking for osmotic pressure.
aaronq
  • aaronq
#1. is asking about osmotic pressure... "1. Determine the osmotic pressure of a solution.."
aaronq
  • aaronq
>_>
anonymous
  • anonymous
Oh yea but still. I said to use PV=nRT for problem 1 meaning temperature would be used. I never mentioned temperature or gas constant for the problem 2 because I forgot to put it there.
anonymous
  • anonymous
thank you guys
anonymous
  • anonymous
No problem.

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