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31 October, 00:50

How many grams of potassium nitrate (KNO3) are required to prepare 0.250L of a 0.700 M

solution (KNO3 = 101.102 g/mol)

O 12.5 g

O 31.2g

25.38

17.78

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Answers (1)
  1. 31 October, 01:03
    0
    17.78g

    Explanation:

    m/M = C*V

    m/101.102 = 0.7 * 0.25

    m = 101.102*0.7*0.25 = 17.78g
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