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13 January, 05:20

A scientist wants to make a solution of tribasic solution phosphate, na3po4, for a laboratory experiment. How many grams of na3po4 will be needed to produce 675 ml of a solution that has a concentration of na + ions of 1.50 m

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  1. 13 January, 06:55
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    55.75g

    Explanation:

    From

    m/M = CV

    Where

    m = required mass of solute

    M = molar mass of solute

    C = concentration of solution

    V = volume of solution=675ml

    Molar mass of solute = 3 (23) + 31 + 4 (16) = 69+31+64=164gmol-1

    Number of moles of sodium ions present = 1.5 * 675/1000 = 1.01 moles

    Since 1 mole of Na3PO4 contains 3 moles of Na+

    It implies that 1.01/3 moles of Na3PO4 are present in solution = 0.34moles

    mass of Na3PO4 = number of moles * molar mass = 0.34 * 164 = 55.75g
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