New RBI Grade B 2025 Notification will come Soon! Enroll Here

    Question

    For preparation of 1000 mL of 0.1 N potassium dichromate

    solution (0.1 N N2Cr2O7, atomic mass of K = 39, Cr = 52, 0 = 16), the amount of analytical grade potassium dichromate required is:
    A 294 gram Correct Answer Incorrect Answer
    B 49.0 gram Correct Answer Incorrect Answer
    C 4.9 gram Correct Answer Incorrect Answer
    D 29.4 gram Correct Answer Incorrect Answer

    Solution

    To calculate the amount of analytical grade potassium dichromate required to prepare a 0.1 N solution, we need to use the following formula: Amount (in grams) = (Normality × Equivalent weight × Volume) / 1000 The equivalent weight of potassium dichromate (K₂Cr₂O_7) can be calculated by considering its molecular weight and the number of equivalents of potassium dichromate present in the formula. The formula weight of K₂Cr₂O_7 is: (2 × atomic mass of K) + (2 × atomic mass of Cr) + (7 × atomic mass of O) = (2 × 39) + (2 × 52) + (7 × 16) = 294 + 104 + 112 = 510 The equivalent weight is then calculated by dividing the formula weight by the number of equivalents. Since potassium dichromate donates two equivalents of Cr₂O_7^2- per mole, the equivalent weight is: Equivalent weight = Formula weight / Number of equivalents = 510 / 2 = 255 g/equivalent Now we can substitute the values into the formula: Amount (in grams) = (0.1 N × 255 g/equivalent × 1000 mL) / 1000 Simplifying further: Amount (in grams) = 0.1 × 255 = 25.5 grams Therefore, the correct answer is option 3: 4.9 grams.

    Practice Next

    Relevant for Exams: