Question

A student plans to prepare 3.0 liters of 0.50 m naoh solution from a 6.0 m naoh stock solution, how many millilters of stock solution should the student use

Answer 1

Answer : The volume of NaOH stock solution used should be, 250 ml

Solution :

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of NaOH solution = 0.50 M

$V_1$ = volume of NaOH solution = 3.0 L

$M_2$ = molarity of NaOH stock solution = 6.0 M

$V_2$ = volume of NaOH stock solution = ?

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

$(0.50M)\times (3.0L)=(6.0M)\times V_2$

$V_2=0.25L=250ml$       (1 L = 1000 ml)

Therefore, the volume of NaOH stock solution used should be, 250 ml