Question

If water is added to 50 ml of a 0.04 M solution so that it fills a 200 ml beaker, what is the final concentration? M

Answer 1

50ML WILL YIELD 0.04M
200ML WILL YIELD xM
50xM=200X0.04
50m=8
M=8/50
m=0.16 moles

Answer 2

Answer : The final concentration is, 0.01 mole/L or 0.01 M

Solution :

According to the neutralization law,

`M_1V_1=M_2V_2`

where,

`M_1` = initial molarity (concentration) of solution = 0.04 M = 0.04 mole/L

`V_1` = initial volume of solution = 50 ml = 0.05 L

`M_2` = final molarity (concentration) of solution = ?

`V_2` = final volume of solution = 200 ml = 0.2 L

Conversion : (1 L = 1000 ml)

Now put all the given values in the above law, we get the volume of NaOH solution.

`(0.04mole/L)* (0.05L)=(M_2)* (0.2L)`

`M_2=0.01mole/L=0.01M`

Therefore, the final concentration is, 0.01 mole/L or 0.01 M