In a reaction involvojg the iodination of acetone, the following volumes were us

Question

In a reaction involvojg the iodination of acetone, the following volumes were used to make up the reaction mixture.

10mL 4.0M acetone + 10mL 1.0M HCl + 10mL .0050 M I2 + 20 mL H20

A. How many moles of acetone were in the mixture? Recall that, for a component A, # moles A=M*V, where M s the molarity of A and V the volume in liters of the solution of A that was used.

B. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50 mL, .050 Lites, and the number of moles of acetone was found in part a.

C. How could you double the molarity of the acetone in the reaction mixture, keeping the total volume at 50 mL and keepin the same concentration of H+ and I2 as in the original mixture?

Explanation / Answer

a)moles of acetone in the mixture= molairty*(volume in L)= 4*10/1000=0.04 moles

b) Molaity of acetone= 0.04/0.050=0.8

c) molarity needs to be doubled, means molairty= 0.8*2= 1.6

the volume of 4M acetone becoomes 20ml.so remove 10 ml water and add 10ml acetone. the mixture volume remains at 50ml.

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