Question

1. You are asked to weigh 250mmol (two hundred fifty millimoles) of a compound with a molecular weight of 80 g/mol. How much of this compound should you weigh?

2. 250 ml of a 3 M stock X is added to 2 L of H_zO in a large flask to make Solution A. You then remove 250 ml of Solution A from the flask. What is the concentration of X now?

Answer 1

1. There is 80 g in 1 mol of the compound.

I mole= 1000 mmol

80 g= 1000 mmol

?= 250 mmol

Grams of compound in 250 mmol= 250/1000 X 80 g= 20 g

20 gram of the compound should be weighed.

2. Stock solution = C1=3M

Initial volume= V1= 250 mL

Final volume= V2= 2L = 2000mL

Final concentration= C2 = ?

C1V1= C2V2

3M X 250 mL=C2 X 2000 mL

Concentration of final solution A= C2= 3M X 250 mL/2000mL= 0.375 M

Now, the 2000 mL solution A has a concentration of 0.375M

0.375 M= 2000 mL

?= 2000 mL-250 mL= 1750 mL

Concentration of X= 1750 mL/2000 mL X 0.375= 0.328 M

Concentration of X in final solution = 0.328 M= 0.33M= 0.33 moles/L