How would you prepare \(60.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HNO}_{3}\) from a stock solution of \(4.00 \mathrm{M} \mathrm{HNO}_{3}\) ?

Short Answer

Expert verified
3.00 mL of the stock solution is needed to prepare 60.0 mL of a 0.200 M HNO3 solution.

Step by step solution

01

Identify the Given Information

The values that have been given are as follows: M1 (initial molarity) = 4.00 M, M2 (final molarity) = 0.200 M, and V2 (final volume) = 60.0 mL. The value that needs to be found is V1 (initial volume).
02

Convert mL to L

Before proceeding, the volumes need to be in the same units. So, V2 is converted from mL to L by division by 1000, yielding V2 = 60.0 mL / 1000 = 0.0600 L.
03

Substitute Values into the Formula

Now substitute the known values into the M1V1 = M2V2 formula, yielding 4.00 M * V1 = 0.200 M * 0.0600 L.
04

Solve for V1

After simplifying, you'll get V1 = (0.200 M * 0.0600 L) / 4.00 M. Thus, V1 = 0.00300 L or 3.00 mL when converted back to mL by multiplying by 1000.

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