Soru
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
Çözüm
3.8
(174 Oylar)
Orçun
Usta · 5 yıl öğretmeni
Uzman doğrulaması
Cevap
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.First, let's calculate the total volume of the mixed solution:Total volume = Volume of HCl solution + Volume of NaCl solutionTotal volume = 200 cm³ + 200 cm³Total volume = 400 cm³Next, let's calculate the total amount of chloride ions contributed by each solution:Amount of chloride ions from HCl = Concentration of HCl × Volume of HCl solutionAmount of chloride ions from HCl = 1 mol/dm³ × 200 cm³Amount of chloride ions from HCl = 200 mmolAmount of chloride ions from NaCl = Concentration of NaCl × Volume of NaCl solutionAmount of chloride ions from NaCl = 5 mol/dm³ × 200 cm³Amount of chloride ions from NaCl = 1000 mmolNow, let's calculate the total amount of chloride ions in the mixed solution:Total amount of chloride ions = Amount of chloride ions from HCl + Amount of chloride ions from NaClTotal amount of chloride ions = 200 mmol + 1000 mmolTotal amount of chloride ions = 1200 mmolFinally, let's calculate the concentration of chloride ions in the mixed solution:Concentration of chloride ions = Total amount of chloride ions / Total volume of the mixed solutionConcentration of chloride ions = 1200 mmol / 400 cm³Concentration of chloride ions = 3 mol/dm³Therefore, the concentration of chloride ions in the mixed solution is 3 mol/dm³.