What is the molality of the 40.0 mass % ethylene glycol 1C2H6O22 solution used for automobile antifreeze?

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Hello everyone. And is probably trying to calculate for the morality of a 32.5% aqueous solution of sucrose. Let's recall what melody is. So morality is equal to the moles of our salutes over the kilograms of our solvent. So we're given the mass percent of 32.5. So this just breaks down to saying that we have 32.5 g of sucrose In 100 g of solution. Alright, so not calculate for the mass Of our solvent. So that's H 20. We just take 100 g and subtracted with 32.5 g, which we get the mass then of 67. g. And of course, we know that we need the kilogram unit. So we'll just do a direct conversion here. So for every one kg we have 1000 g. We see here that the units of grounds will cancel Leaving us with the mass of HO being 0.0675 kg. Looking for the moles of solute, which is sucrose. So we're given that we have 32.5 g of sucrose and then we can go ahead and just use the molar mass here of super gross. So that's just one mole For every 342.3 g of sucrose. So we see here that the units of grams, low mass will cancel leaving us with just the units of moles and the numerical value is 0. moles. So I'm putting this into this equation here. Then we can get that. The morality is equal to 0.094946 moles, And then denominator is just kg, and that is 0.0675 kg. So once you put that into a calculator, you get the value of 1.41 and of course, our lower case M for units of morality. And this right here is going to be my final answer for this problem.

What is the molality of the 40.0 mass % ethylene glycol 1C2H6O22 solution used for automobile antifreeze?

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Key Concepts

Molality

Mass Percent

Molar Mass