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8 July, 15:53

a scientist uses 68 grams of CaCo3 to prepare 1.5 liters of solution. what is the molarity of this solution?

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  1. 8 July, 19:02
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    Answer: 0.4533mol/L

    Explanation:

    Molar Mass of CaCO3 = 40+12 + (16x3) = 40+12+48 = 100g/mol

    68g of CaCO3 dissolves in 1.5L of solution.

    Xg of CaCO3 will dissolve in 1L i. e

    Xg of CaCO3 = 68/1.5 = 45.33g/L

    Molarity = Mass conc. (g/L) / molar Mass

    Molarity = 45.33/100 = 0.4533mol/L
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