Question

a chemist prepares a solution of aluminum chloride (alcl₃) by measuring out 4.7 µmol of aluminum chloride into a 350. ml volumetric flask and filling the flask to the mark with water. calculate the concentration in mol/l of the chemists aluminum chloride solution. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Recall the formula for molarity

The formula for molarity \( M=\frac{n}{V} \), where \( n \) is the number of moles and \( V \) is the volume in liters.

Step2: Convert volume from mL to L

Given \( V = 350\space mL\). Since \( 1\space L=1000\space mL\), then \( V=\frac{350}{1000}=0.350\space L\).

Step3: Substitute values into the formula

We have \( n = 4.7\times10^{-4}\space mol\) and \( V = 0.350\space L\). Then \( M=\frac{4.7\times 10^{-4}}{0.350}\).

Step4: Calculate the molarity

\( M=\frac{4.7\times 10^{-4}}{0.350}\approx1.34286\times 10^{-3}\). Rounding to two significant figures (since \( 4.7\) has two significant figures), \( M = 1.3\times 10^{-3}\space M\).

Answer:

\( 1.3\times 10^{-3}\space M\)