how many grams of methanol are contained in 0.100 L of 1.71 M aqueous methanol?

Question

Question

Answer 1

Answer #1

0.100 L of 1.71 M aqueous methanol corresponds to

$1.71 \: mol/L \times 0.100 \: L = 0.171 \: moles$ of methanol.

The molecular mass of methanol is 32 g/mol.

The number of grams of methanol $0.171 \: moles \times 32 \: g/mol = 5.47 \: g$

Answer 2

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