Question

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

Answer 1

b . 0.351 L.

Step-by-step 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=(C_2V_2)/(C_1)`

Now, by plugging in the given data we obtain:

`V_1=(1.2M*0.585L)/(2.0M)\\\\V_1=0.351L`

Therefore, the answer is b . 0.351 L.

Best regards!