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Some fertilizer blends contain magnesium nitrate [tex]\left( Mg \left( NO _3\right)_2\right)[/tex]. Suppose that a chemist has 1.24 liters of a 2.13 M solution of magnesium nitrate. If the chemist dilutes the solution to 1.60 M, what is the volume of the new solution?

Express your answer to three significant figures.

The volume of the new solution is [tex]$\square$[/tex] liters.


Sagot :

To solve the problem of finding the volume of the new solution when diluting a chemist's magnesium nitrate solution, we'll use the principle of conservation of moles during dilution. This is encapsulated in the formula:

[tex]\[ C_1 V_1 = C_2 V_2 \][/tex]

where:
- [tex]\( C_1 \)[/tex] is the initial concentration
- [tex]\( V_1 \)[/tex] is the initial volume
- [tex]\( C_2 \)[/tex] is the final concentration
- [tex]\( V_2 \)[/tex] is the final volume

Let's break it down step-by-step:

1. Identify initial parameters:
- [tex]\( C_1 = 2.13 \)[/tex] M (initial concentration)
- [tex]\( V_1 = 1.24 \)[/tex] liters (initial volume)

2. Identify the final concentration:
- [tex]\( C_2 = 1.60 \)[/tex] M (final concentration)

3. Rearrange the formula to solve for [tex]\( V_2 \)[/tex]:

[tex]\[ V_2 = \frac{C_1 V_1}{C_2} \][/tex]

4. Plug in the known values:

[tex]\[ V_2 = \frac{(2.13 \, \text{M}) \cdot (1.24 \, \text{L})}{1.60 \, \text{M}} \][/tex]

5. Calculate the final volume:

[tex]\[ V_2 = \frac{2.6412 \, \text{M} \cdot \text{L}}{1.60 \, \text{M}} \][/tex]
[tex]\[ V_2 = 1.65075 \, \text{L} \][/tex]

6. Round to three significant figures:

[tex]\[ V_2 \approx 1.65 \, \text{L} \][/tex]

Thus, the volume of the new solution is [tex]\( \boxed{1.65} \)[/tex] liters.