Question

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 1

Answer: 0.52 L of 15 M $NH_3(aq)$ 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.

$C_1V_1(stock)=C_2V_2$ (to be prepared)

where,

$C_1$ =concentration of stock solution = 15 M

$V_1$ = volume of stock solution = ?

$C_2$ = concentration of solution to be prepared = 0.52 M

$V_2$ = volume of solution to be prepared = 15 L

$15\times V_1=0.52\times 15$

$V_1=0.52L$

Thus 0.52 L of 15 M $NH_3(aq)$ will be used to prepare this amount of 0.52 M base