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1 September, 13:47

How many grams of methanol (CH 3 OH, FM 32.04) are contained in 0.100 L of 1.71 M aqueous methanol (i. e., 1.71 mol CH 3 OH/L solution) ?

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  1. 1 September, 15:23
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    5.48 g

    Explanation:

    The concentration in M represents the mol/L, or the number of moles in each L of the solution, and it can be calculated by the number of moles (n) divided by the volume of the solution (V) in L, so:

    M = n/V

    1.71 = n/0.100

    n = 0.171 mol

    The molar mass of methanol is 32.04 g/mol, and it is the mass (m) divided by the number of moles:

    FM = m/n

    32.04 = m/0.171

    m = 5.48 g
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