What is the molarity of a solution prepared from 32.0 grams of methanol (ch3oh, density = 0.792 g/ml) with 125 milliliters of ethanol (ch3ch2oh)? assume the volumes are additive?

Molarity = moles of solute/ volume of solution

Mass of methanol = 32.0 grams

Molar mass of methanol = 32.04 g/mol

Moles of methanol (solute) = mass / molar mass = 32.0 grams/ 32 g/mol

Moles of methanol (solute) = 1 mole

Density of methanol = 0.792 g/ml

Volume of methanol = mass / density = 32.0 g/ 0.792 g/ml = 40.4 ml

Volume of ethanol = 125 ml

Volume of solution = Volume of methanol (solute)  +  Volume of ethanol (solvent)

Volume of solution = 40.4 ml + 125 ml = 165.4 ml = 0.1654 L

Conversion factor: 1L = 1000ml

Molarity = 1 / 0.1654 = 6.046 M

So the molarity of the solution is 6.046 M

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explanation:

the answer is na (sodium)