\( 15 \mathrm{~cm}^{\frac{3}{3}} \) of a solution of sodium chloride saturated at \( 20^{\circ} \mathrm{C} \) required \( 35.5 \mathrm{~cm}^{3} \) of silver trioxonitrate \( (\mathrm{V}) \) solution containing 240 gdm for complete precipitation. Calculate the solubility of sodium chloride in moldm \( { }^{-3}(\mathrm{~N}=14 ; \mathrm{O}=16 ; \mathrm{Ag}=108) \)

To calculate the solubility of sodium chloride (NaCl) in moldm⁻³, we first need to determine the amount of NaCl that was present in the 15 cm³ of solution by considering the silver nitrate (AgNO₃) that reacted with it. The reaction that takes place is: \[ \text{NaCl} + \text{AgNO}_3 \rightarrow \text{NaNO}_3 + \text{AgCl} \] From the problem, we know that 35.5 cm³ of silver trioxonitrate (AgNO₃) was used, with a concentration of 240 gdm⁻³. To find the number of moles of AgNO₃ used, we convert cm³ to dm³ (because molarity is often expressed in grams per dm³): \[ 35.5 \, \text{cm}^3 = 0.0355 \, \text{dm}^3 \] Now, we can calculate the moles of AgNO₃ used: \[ \text{Moles of AgNO}_3 = \text{Concentration} \times \text{Volume} = 240 \, \text{gdm}^{-3} \times 0.0355 \, \text{dm}^3 = 8.52 \, \text{g} \] Next, we find out how many moles of AgNO₃ this mass corresponds to: Molar mass of AgNO₃ = 108 (Ag) + 14 (N) + 48 (O) = 170 g/mol. \[ \text{Moles of AgNO}_3 = \frac{8.52 \, \text{g}}{170 \, \text{g/mol}} = 0.0501 \, \text{mol} \] From the balanced equation, it’s clear that 1 mole of NaCl will produce 1 mole of AgCl. Therefore, the moles of NaCl in the saturated solution is also 0.0501 mol. To determine the solubility in moldm⁻³, we divide the moles of NaCl (0.0501 mol) by the volume of the solution in dm³ (15 cm³ = 0.015 dm³): \[ \text{Solubility of NaCl} = \frac{0.0501 \, \text{mol}}{0.015 \, \text{dm}^3} \approx 3.34 \, \text{mol/dm}^3 \] Thus, the solubility of sodium chloride in the solution is approximately **3.34 moldm⁻³**.