Problem A solution is prepared by dissolving 5.00 g of potassium sulfate (K2SO4) in 100 mL of water. The van't Hoff factor for K2SO4 is 3. The experimenter decided to dilute the solution with another 100 mL of water. 1) What two particles will potassium sulfate dissociate into? *

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Chapter1: Chemical Foundations
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SUBJECT : GENERAL CHEMISTRY 2

TOPIC : OSMOTIC PRESSURE 

NO NEED FOR SOLUTION JUST THE ANSWER THANK YOU GOODLUCK : ) I'LL WAIT FOR YOUR ANSWER 

Refer to the given below:
Important: Assume that the volume of the solution is equal to the total volume of the
solvent.
Problem
A solution is prepared by dissolving 5.00 g of potassium sulfate (K₂SO4) in 100 mL of
water. The van't Hoff factor for K2SO4 is 3. The experimenter decided to dilute the
solution with another 100 mL of water.
1) What two particles will potassium sulfate dissociate into? *
Your answer
Transcribed Image Text:Refer to the given below: Important: Assume that the volume of the solution is equal to the total volume of the solvent. Problem A solution is prepared by dissolving 5.00 g of potassium sulfate (K₂SO4) in 100 mL of water. The van't Hoff factor for K2SO4 is 3. The experimenter decided to dilute the solution with another 100 mL of water. 1) What two particles will potassium sulfate dissociate into? * Your answer
4) What is the molarity of this solution? *
Your answer
5) If the temperature is at 25 degrees celsius. What is this temperature in *
kelvin?
Your answer
6) What is the final osmotic pressure of the solution? *
π = iMRT
= osmotic pressure
i = van't Hoff factor
w
M = molarity of the solution
R = gas constant (0.082 L atm /mol K
T = temperature in kelvin
Your answer
Transcribed Image Text:4) What is the molarity of this solution? * Your answer 5) If the temperature is at 25 degrees celsius. What is this temperature in * kelvin? Your answer 6) What is the final osmotic pressure of the solution? * π = iMRT = osmotic pressure i = van't Hoff factor w M = molarity of the solution R = gas constant (0.082 L atm /mol K T = temperature in kelvin Your answer
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