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19 April, 10:57

How many grams of KBr are required to make 350. mL of a 0.115 M KBr solution?

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  1. 19 April, 11:07
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    0.115 M means that 0.115 moles of KBr are contained in a volume of 1000 ml, therefore a volume of 350 ml will have (0.115 * 0.35) = 04025 moles

    From the formula of molarity moles = molarity * volume in liters

    1 mole of KBr is equivalent to 119 g

    Therefore, the mass = 0.04025 * 119 g = 4.79 g
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