Question

how would you prepare 70.0 ml of 0.200 m hno3 from a stock solution of 4.00 m hno3? what volume (in ml) of the 4.00 m hno3 solution will you need? round your answer to 3 significant digits.

Explanation:

Step1: Use dilution formula

The dilution formula is \(C_1V_1 = C_2V_2\), where \(C_1\) is the initial concentration, \(V_1\) is the initial volume, \(C_2\) is the final concentration, and \(V_2\) is the final volume.
We are given \(C_1 = 4.00\space M\), \(C_2=0.200\space M\), and \(V_2 = 70.0\space mL\). We need to find \(V_1\).

Step2: Rearrange the formula to solve for \(V_1\)

From \(C_1V_1 = C_2V_2\), we can solve for \(V_1\) as \(V_1=\frac{C_2V_2}{C_1}\)
Substitute the given values: \(V_1=\frac{0.200\space M\times70.0\space mL}{4.00\space M}\)

Step3: Calculate \(V_1\)

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Answer:

\(3.50\space mL\)