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An experiment requires 0.52 M NH3(aq). The stockroom manager estimates that 15 L of the base is needed. What volume of 15 M NH3(aq) will be used to prepare this amount of 0.52 M base

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

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