What is the molarity of a solution prepared by dissolving 5.85 g of NaCl in 500 mL of water? (Molar mass of NaCl = 58.5 g/mol)
The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL of 0.010 M barium nitrate with 100 mL of 0.10 M sodium iodate is $X \times 10^{-6} \, \text{mol dm}^{-3}$. The value of $X$ is ------.Use: Solubility product constant $(K_{sp})$ of barium iodate = $1.58 \times 10^{-9}$