How much 2.0 M NH4NO3 is needed to make 0.585 L of 1.2 M NH4NO3 solution? a
0.176 L
b
0.351 L
c
0.301 L
d
0.140 L
e
0.234 L

b . 0.351 L.

Explanation:

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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:

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

Now, by plugging in the given data we obtain:

Therefore, the answer is b . 0.351 L.

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