Question

A student requires 2.00 L of 0.100 M NH4NO3 from a 1.75 M NH4NO3 stock solution. What is the correct way to get the solution?

Answer 1

Hello!

A student requires 2.00 L of 0.100 M NH4NO3 from a 1.75 M NH4NO3 stock solution. What is the correct way to get the solution ?

We have the following data:

M1 (initial molarity) = 0.100 M (or mol/L)

V1 (initial volume) = 2.00 L

M2 (final molarity) = 1.75 M (or mol/L)

V2 (final volume) = ? (in L)

Let's use the formula of dilution and molarity, so we have:

$M_{1} * V_{1} = M_{2} * V_{2}$

$0.100 * 2.00 = 1.75 * V_{2}$

$0.2 = 1.75\:V_2$

$1.75\:V_2 = 0.2$

$V_2 = \dfrac{0.2}{1.75}$

$\boxed{\boxed{V_2 \approx 0.114\:L}}\:\:or\:\:\:\boxed{\boxed{V_2 \approx 114\:mL}}\:\:\:\:\:\bf\green{\checkmark}$

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Answer 2

The student needs to dilute the stock to get the proper concentration. And during the dilution, the molar number of solute will not change. So the volume of stock needed is 2*0.1/1.75=0.114 L. Then dilute to 2.00 L.