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29 July, 20:13

What mass, in grams, of aluminum hydroxide will be required to prepare the 4 L of a 1.75 M solution?

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  1. 29 July, 20:26
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    Answer: 0.53g

    Explanation:

    No of moles = volume * molarity/1000

    We have the volume and the molarity

    Volume=4L

    Molarity=1.7M

    No of moles = 4*1.7/1000

    No of moles = 0.0068moles

    Remember also that

    No of moles = mass given/molar mass

    Molar mass of Al (OH) 3

    Al = 27

    O=16

    H=1

    Molar mass = Al + (O+H) 3

    Molar mass = 27 + (16+1) 3

    Molar mass = 27 + (17) 3

    Molar mass = 27+51

    Molar mass = 78g/mol

    To get the mass

    Mass given = no of moles * molar mass

    Mass = 0.0068*78

    Mass = 0.53g
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