a. The nonvolatile, nonelectrolyte chlorophyll , C 55 H 72 MgN 4 O 5 ( 893.50 g/

Question

a. The nonvolatile, nonelectrolyte chlorophyll, C55H72MgN4O5 (893.50 g/mol), is soluble in ethanol CH3CH2OH.  

Calculate the osmotic pressure generated when 10.4 grams of chlorophyll are dissolved in 184 ml of a ethanol solution at 298 K.  

The molarity of the solution is _____ M.

The osmotic pressure of the solution is _______atmospheres.

b. The nonvolatile, nonelectrolyte chlorophyll ,  C55H72MgN4O5 (893.5 g/mol), is soluble in diethyl ether CH3CH2OCH2CH3.  

How many grams of chlorophyll are needed to generate an osmotic pressure of 1.21 atm when dissolved in 236 ml of a diethyl ethersolution at 298 K.

_______ grams chlorophyll

Explanation / Answer

1) osmotic pressure=P=cRT c=conc of solution,

C=molarity=moles/volume

Moles=10.4g/(893.50 g/mol)=0.01164 moles

Molarity=0.01164 moles/184 ml=0.01164 mol/0.184 L=0.0632 M

P=0.0632 mol/L * 0.0821 L atm/K mol * 298K=1.548 atm

2)Let x grams of chlorophyll are needed to generate an osmotic pressure of 1.21 atm

Molarity=c=moles/volume

Moles=x/(893.50 g/mol)

Molarity= x/(893.50 g/mol)*1/0.236 L=x/210.866 M

osmotic pressure=P=cRT = x/210.866 mol/L RT

1.21 atm= x/210.866 mol/L * 0.0821 L atm/K mol*298K

X=9.653 g

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