Answer :
Answer: 0.52 L of 15 M [tex]NH_3(aq)[/tex] will be used to prepare this amount of 0.52 M base.
Explanation:
But on diluting the number of moles remain same and thus we can use molarity equation.
[tex]C_1V_1(stock)=C_2V_2[/tex] (to be prepared)
where,
[tex]C_1[/tex] =concentration of stock solution = 15 M
[tex]V_1[/tex] = volume of stock solution = ?
[tex]C_2[/tex] = concentration of solution to be prepared = 0.52 M
[tex]V_2[/tex] = volume of solution to be prepared = 15 L
[tex]15\times V_1=0.52\times 15[/tex]
[tex]V_1=0.52L[/tex]
Thus 0.52 L of 15 M [tex]NH_3(aq)[/tex] will be used to prepare this amount of 0.52 M base