Question

How much 2.0 M NH4NO3 is needed to make 0.585 L of 1.2 M NH4NO3 solution?

Answer 1

Answer:

b . 0.351 L.

Explanation:

Hello!

In this case, since diluted solutions are prepared by adding an extra amount of diluent to a stock-concentrated solution, we infer that the number of moles of solute remains the same, therefore we can write:

$C_1V_1=C_2V_2$

Thus, solving for the volume of the stock solution, V1, we obtain:

$V_1=\frac{C_2V_2}{C_1}$

Now, by plugging in the given data we obtain:

$V_1=\frac{1.2M*0.585L}{2.0M}\\\\V_1=0.351L$

Therefore, the answer is b . 0.351 L.

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