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 : The volume of [tex]NH_4NO_3[/tex] stock solution is, 0.114 L
Solution :
According to the dilution law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of [tex]NH_4NO_3[/tex] solution = 0.100 M
[tex]V_1[/tex] = volume of [tex]NH_4NO_3[/tex] solution = 2.00 L
[tex]M_2[/tex] = molarity of [tex]NH_4NO_3[/tex] stock solution = 1.75 M
[tex]V_2[/tex] = volume of [tex]NH_4NO_3[/tex] stock solution = ?
Now put all the given values in the above law, we get the volume of [tex]NH_4NO_3[/tex] stock solution.
[tex](0.100M)\times (2.00L)=(1.75M)\times V_2[/tex]
[tex]V_2=0.114L[/tex]
Therefore, the volume of [tex]NH_4NO_3[/tex] stock solution is, 0.114 L
