40 - (CHIMM/12 ML. Sommer 3022.01) - Stokichiomeric Relationship What is the concentration of chloride ions, in moldm^-3 in a solution formed by mixing 200cm^3 of 1moldm^-1 HCl with 200cm^2 of 5moldm^-1NaCl B. a C. 3 D. s

To find the concentration of chloride ions in the solution formed by mixing the two solutions, we need to consider the total volume of the mixed solution and the total amount of chloride ions contributed by each solution.<br /><br />First, let's calculate the total volume of the mixed solution:<br />Total volume = Volume of HCl solution + Volume of NaCl solution<br />Total volume = 200 cm³ + 200 cm³<br />Total volume = 400 cm³<br /><br />Next, let's calculate the total amount of chloride ions contributed by each solution:<br />Amount of chloride ions from HCl = Concentration of HCl × Volume of HCl solution<br />Amount of chloride ions from HCl = 1 mol/dm³ × 200 cm³<br />Amount of chloride ions from HCl = 200 mmol<br /><br />Amount of chloride ions from NaCl = Concentration of NaCl × Volume of NaCl solution<br />Amount of chloride ions from NaCl = 5 mol/dm³ × 200 cm³<br />Amount of chloride ions from NaCl = 1000 mmol<br /><br />Now, let's calculate the total amount of chloride ions in the mixed solution:<br />Total amount of chloride ions = Amount of chloride ions from HCl + Amount of chloride ions from NaCl<br />Total amount of chloride ions = 200 mmol + 1000 mmol<br />Total amount of chloride ions = 1200 mmol<br /><br />Finally, let's calculate the concentration of chloride ions in the mixed solution:<br />Concentration of chloride ions = Total amount of chloride ions / Total volume of the mixed solution<br />Concentration of chloride ions = 1200 mmol / 400 cm³<br />Concentration of chloride ions = 3 mol/dm³<br /><br />Therefore, the concentration of chloride ions in the mixed solution is 3 mol/dm³.