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Step 1: Calculate the number of moles of solute (Na₂CO₃).
The molar mass of Na₂CO₃ is 106 g/mol.
[tex]\begin{aligned} n_{\text{solute}} & = \frac{mass_{\text{solute}}}{MM_{\text{solute}}} \\ & = \frac{\text{5.3 g}}{\text{106 g/mol}} \\ & = \text{0.050 mol} \end{aligned}[/tex]
Step 2: Calculate the molarity of solution.
The volume of solution must be expressed in liters. 100 mL = 0.10 L
[tex]\begin{aligned} M & = \frac{n_{\text{solute}}}{V_{\text{solution}}} \\ & = \frac{\text{0.050 mol}}{\text{0.10 L}} \\ & = \text{0.50 mol/L} \\ & = \boxed{0.50 \: M} \end{aligned}[/tex]
Hence, the molarity of 5.3 g anhydrous sodium carbonate dissolved in 100 mL of solution is 0.50 M.
[tex]\\[/tex]
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