This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
Hey Gobte, good to see you again. Here are the calculations based on the previous experiment involving the standardization of silver nitrate solution using potassium chloride. 1) To calculate the concentration of the KCl solution. Step 1: Calculate the moles of KCl. From the previous problem's Table 1, the mass of KCl used was 1.86 g. The relative molecular mass of KCl is 74.6 g/mol. Moles of KCl = MassMolar mass = 1.86 g74.6 g/mol ≈ 0.024932976 mol Step 2: Determine the volume of the KCl solution. Assuming the KCl was dissolved in 250 cm^3 to prepare the standard solution (a common practice for such preparations). Volume of KCl solution = 250 cm^3 = 0.250 dm^3. Step 3: Calculate the concentration of KCl. Concentration of KCl = Moles of KClVolume of solution (dm^3) Concentration of KCl = 0.024932976 mol0.250 dm^3 ≈ 0.0997319 mol/dm^3 Rounding to three significant figures: Concentration of KCl = 0.0997 mol/dm^3 2) To determine the concentration of the AgNO_3 solution. Step 1: Identify the known values and the balanced chemical equation. The balanced chemical equation for the reaction between AgNO_3 and KCl is: AgNO_3(aq) + KCl(aq) AgCl(s) + KNO_3(aq) The mole ratio of AgNO_3 to KCl is 1:1. So, n_AgNO_3 = 1 and n_KCl = 1. From the previous problem's context and calculations: Concentration of KCl (M_KCl) = 0.0997319 mol/dm^3 (unrounded value from Q1) Volume of KCl (V_KCl) = 25 cm^3 (pipetted into the conical flask) Mean titre volume of AgNO_3 (V_AgNO_3) = 24.95 cm^3 (calculated in the previous step for the last question you sent). Step 2: Use the titration formula. M_AgNO_3 V_AgNO_3n_AgNO_3 = M_KCl V_KCln_KCl Rearranging to solve for M_AgNO_3: M_AgNO_3 = M_KCl V_KCl n_AgNO_3V_AgNO_3 n_KCl Step 3: Substitute the values and calculate M_AgNO_3. M_AgNO_3 = (0.0997319 mol/dm^3) × (25 cm^3) × 1(24.95 cm^3) × 1 M_AgNO_3 = (2.4932975)/(24.95) mol/dm^3 M_AgNO_3 ≈ 0.09993176 mol/dm^3 Rounding to three significant figures: Concentration of AgNO_3 = 0.0999 mol/dm^3 What's next?