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The amount (in grams) of potassium dichromate (MW = 294) present in 75 mL of 0.16 M aqueous solution is _______
    Correct answer is between '3.52,3.54'. Can you explain this answer?
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    The amount (in grams) of potassium dichromate (MW = 294) present in 75...
    Calculation of Amount of Potassium Dichromate in Solution

    Given:
    - Volume of solution = 75 mL
    - Concentration of solution = 0.16 M
    - Molecular weight of potassium dichromate = 294 g/mol

    Formula:
    - Amount of solute (in grams) = Concentration × Volume × Molecular weight

    Solution:

    - Convert volume from milliliters to liters:
    - 75 mL = 0.075 L

    - Substitute given values into the formula:
    - Amount of solute (in grams) = 0.16 M × 0.075 L × 294 g/mol
    - Amount of solute (in grams) = 3.528 g

    - Round off the answer to two decimal places:
    - Amount of potassium dichromate present in 75 mL of 0.16 M aqueous solution is 3.54 g.

    Therefore, the correct answer is between 3.52 and 3.54 g.
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    Community Answer
    The amount (in grams) of potassium dichromate (MW = 294) present in 75...
    No.of .moles in 0.075L of k2cr2o7 = molarity ×volume
    = 0.16 × 0.075
    = 0.012

    =no.of.moles = wt in g/molecular weight
    =0.16 =wt in g /294
    wt in g = 0.16×294= 3.528


    answer(3.528)
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    The amount (in grams) of potassium dichromate (MW = 294) present in 75 mL of 0.16 M aqueous solution is _______Correct answer is between '3.52,3.54'. Can you explain this answer?
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