Question

what is the molarity of lead in this sample obtained from aliyas house? report your answer to 2 significant figures. molarity (m) = moles of solute / liters of solution molarity is the amount of moles of a substance dissolved in a liter of solution (usually a water - based solution). lead ii carbonate pbco₃ 267.21 grams per mol 1000 ml = 1l solute:.0000028 grams lead (ii) carbonate solution: 100 ml solution

Explanation:

Step1: Convert grams to moles

Moles of \(PbCO_3=\frac{mass}{molar\ mass}\). Given mass of \(PbCO_3 = 0.0000028\ g\) and molar mass of \(PbCO_3=267.21\ g/mol\).

$$n=\frac{0.0000028\ g}{267.21\ g/mol}\approx1.05\times10^{-8}\ mol$$

Step2: Convert volume from mL to L

Given volume \(V = 100\ mL\). Since \(1000\ mL=1\ L\), then \(V=\frac{100}{1000}\ L = 0.1\ L\)

Step3: Calculate molarity

Using the formula \(M=\frac{n}{V}\), where \(n\) is moles of solute and \(V\) is liters of solution.

$$M=\frac{1.05\times 10^{-8}\ mol}{0.1\ L}=1.05\times10^{-7}\ M\approx1.1\times 10^{-7}\ M$$

Answer:

\(1.1\times 10^{-7}\)