Question
Asked Aug 28, 2019
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How many grams of methanol (CH3OH, FM 32.04) are contained in 0.440 L of 1.63 M aqueous methanol (i.e., 1.63 mol CH3OH/L solution)?

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Expert Answer

Step 1

Step 1-

We have to calculate the mass of methanol in grams that are contained in 0.440 L of 1.63 M aqueous methanol.

Thus, V = 0.440 L

Molarity = 1.63 M

Molar mass of methanol = 32.04 g/mol

Step 2

Step 2-

Molarity is a term used to express the concentration of a solution and is defined as the number of moles of solute per litre of solution.

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Molarity =Number of moles of solute Volume of solution in L

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Step 3

Step 3-Calculate the moles of methanol...

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Number of moles of methanol = Molarity x Volume of solution in L = (1.63 M) x (0.440 L) 0.720 moles

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