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14 November, 14:26

How many grams of cabr2 are needed to prepare 4.65 l of a 7.65 m solution?

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  1. 14 November, 16:18
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    Molarity is defined as the number of moles of solute dissolved in 1 L of solution

    the molarity of the solution to be prepared is 7.65 M

    the volume to be prepared of CaBr₂ is 4.65 L

    the number of CaBr₂ moles are - 7.65 mol/L x 4.65 L = 35.6 mol

    the molar mass of CaBr₂ - 200 g/mol

    mass of 35.6 mol - 35.6 mol x 200 g/mol = 7120 g

    mass required of CaBr₂ to make the solution is 7120 g
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