Question

a chemist must dilute 83.1 ml of 5.24 m aqueous iron(ii) bromide (febr₂) solution until the concentration falls to 2.00 m. shell do this by adding distilled water to the solution until it reaches a certain final volume. calculate this final volume, in liters. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Use the dilution formula

The dilution formula is \(C_1V_1 = C_2V_2\), where \(C_1 = 5.24\space M\), \(V_1=83.1\space mL\), and \(C_2 = 2.00\space M\). We need to find \(V_2\).

$$V_2=\frac{C_1V_1}{C_2}$$

Step2: Substitute the values

Substitute \(C_1 = 5.24\), \(V_1 = 83.1\), and \(C_2=2.00\) into the formula.

$$V_2=\frac{5.24\times83.1}{2.00}$$

$$V_2=\frac{435.444}{2.00}=217.722\space mL$$

Step3: Convert milliliters to liters

Since \(1\space L = 1000\space mL\), then \(V_2=\frac{217.722}{1000}\space L\)

$$V_2 = 0.217722\space L$$

Step4: Round to the correct number of significant digits

The least number of significant digits in the given values (\(C_1 = 5.24\) has three, \(V_1 = 83.1\) has three, \(C_2 = 2.00\) has three) is three. So, \(V_2\approx0.218\space L\)

Answer:

\(0.218\space L\)