Question

A student requires 2.00 L of 0.100 M NH4NO3 from a 1.75 M NH4NO3 stock solution. What is the correct way to get the solution?

Answer 1

To solve this we use the equation,

 

M1V1 = M2V2

 

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

 

1.75 M x V1 = 0.100 M x 2.0 L

V1 = -.11 L

Answer 2

Answer : The volume of $NH_4NO_3$ stock solution is, 0.114 L

Solution :

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of $NH_4NO_3$ solution = 0.100 M

$V_1$ = volume of $NH_4NO_3$ solution = 2.00 L

$M_2$ = molarity of $NH_4NO_3$ stock solution = 1.75 M

$V_2$ = volume of $NH_4NO_3$ stock solution = ?

Now put all the given values in the above law, we get the volume of $NH_4NO_3$ stock solution.

$(0.100M)\times (2.00L)=(1.75M)\times V_2$

$V_2=0.114L$

Therefore, the volume of $NH_4NO_3$ stock solution is, 0.114 L