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5 December, 03:28

How many atoms are in 36.1 g of boron

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  1. 5 December, 04:01
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    The answer is 2.011 * 10^24

    Explanation:

    Looking for atoms in boron,

    Molar mass of B = 10.81g/mol

    There is 36.1g of B,.

    Using avogadro constant = 6.023 * 10^23

    36.1g of B = 6.023 * 10^23/10.81 * 36.1g

    That is, 6.023*10^23 divided by 10.81 multiplied by 36.1g

    From this calculation, we get

    36.1g of boron = 2.011 * 10^24atoms.
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