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) ?

Question

Dennis Blankenship

Chemistry

20 October, 20:24

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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Answers (1)

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Makhi York · Answer 1

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