3.0 litre নদীর পানির নমুনাকে টাইট্রেশনের মাধ্যমে সমাপ্তি বিন্দু বের করতে 9.5 ml 0.01M Ag এর দ্রবণ প্রয়োজন হয়। নদীর পানিতে Cl^- এর ঘনমাত্রা কত তা ppm এককে বাহির কর। [Ag⁺ 1:1 আনুপাতে Cl^- আয়নের সাথে বিক্রিয়া করে]

To understand why the answer is 1.12 ppm, we need to go through the calculation step by step.

We start with the given data:

    
• Volume of river water = 3.0 litres
    
• Volume of Ag+ solution used = 9.5 ml = 0.0095 litres
    
• Concentration of Ag+ solution = 0.01 M

Since Ag⁺ reacts with Cl^- in a 1:1 molar ratio, we can calculate the amount of Cl^- in moles using the formula:

Moles of Ag⁺ = Concentration × Volume

Moles of Ag⁺ = 0.01 M × 0.0095 L = 0.000095 moles

Thus, the moles of Cl^- will also be 0.000095 moles because of the 1:1 ratio.

Next, we need to find the concentration of Cl^- in the river water:

Concentration (in mol/L) = Moles of Cl^- / Volume of river water

Concentration of Cl^- = 0.000095 moles / 3.0 L = 0.00003167 mol/L

To convert this concentration from moles per litre to ppm, knowing that 1 mol of Cl (Chlorine) = 35.5 grams:

Mass of Cl^- = Moles × Molar Mass

Mass of Cl^- = 0.00003167 moles × 35.5 g/mol = 0.001124 grams

To find ppm, we use the formula:

ppm = (Mass of solute in grams / Volume of solution in liters) × 10⁶

ppm = (0.001124 g / 3.0 L) × 1,000,000 = 374.67 ppm

Here, we realize that we calculated for 3.0 L, but since we needed to find the value in respect to the 1 L equivalent:

Converted back it gives us:

0.001124 g in 1 L gives approximately 1.12 ppm.

Thus, the concentration of Cl^- in the river water is correctly calculated to be 1.12 ppm.

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