What is the molality of a solution of 31.8 g of ethanol (CH₃CH₂OH) in 405 mL of water if the density of water is 1.0 g/mL? A) 78.5 m B) 1.58 m C) 1.70 m D) 0.0298 m E) 0.0307 m

Question

Quizplus · Accepted Answer

First, we need to convert the volume of water to mass:

Density of water = 1.0 g/mL
Volume of water = 405 mL

Mass of water = density x volume = 1.0 g/mL x 405 mL = 405 g

Next, we can calculate the moles of ethanol:

Mass of ethanol = 31.8 g
Molar mass of ethanol = 46.07 g/mol

Moles of ethanol = mass/molar mass = 31.8 g/46.07 g/mol = 0.689 mol

Finally, we can calculate the molality:

Molality = moles solute/kg solvent

Since we have 405 g of water, which is equivalent to 0.405 kg:

Molality = 0.689 mol/0.405 kg = 1.70 m

Therefore, the answer is C, 1.70 m.

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