Supplements and Herbs

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Butylene Glycol

Question: What mass of butylene glycol, HOCH2CH2CH2CH2OH, is required to lower the freezing point of 90.0 mL of water? What mass of butylene glycol, HOCH2CH2CH2CH2OH, is required to lower the freezing point of 90.0 mL of water to -4.00oC? Use d = 1.00 g/mL for water. a- 2.15 g b- 3.77 g c- 17.4 g d- 63.4 g

Answer: Use the freezing point depression formula: ΔT = Kf * m * i. ΔT is difference in temp. between pure solvent (water) freezing pt. and the freezing point of the solution. Kf is freezing point depression constant (the Kf for water is 1.86 *C/m). m is concentration in units of molality i is the van't hoff factor - the number of particles resulting from the given solution (since butylene glycol is mostly a carbon chain, it won't dissociate at all into ions when dissolved in water, so i = 1). First find the value for ΔT, it should be a positive number. It's the temp difference between the freezing pt. of pure water (which is 0 *C) and the freezing pt. of the solution. So ΔT = 0 *C - (-4.00 *C) = 4 *C. Now you know everything except for the molality. Molality is defined as moles solute / kg solvent. You don't know how many moles of solute you have, BUT you can solve for it. Your solvent is water, and you have 90 mL of it. This is the same thing as 90 g of water (b/c the density of water is 1.00 g/mL). 1000 g = 1 kg, so 90 g = 0.09 kg. Now plug in what you know, and solve for 'x' moles of butylene glycol you have. ΔT = Kf * m * i 4 *C = (1.86 *C/m) * (x moles / 0.09 kg) * (1) After a little algebra, I get x to be 0.1935 moles of butylene glycol. BUT the problem asks for a mass of butylene glycol, so all you need to do is convert the moles you have to grams. The molar mass of butylene glycol is about 90.12 g/mol, so the grams of butylene glycol would be: 0.1935 moles * (90.12 g / 1 mol) = 17.44 grams, which is closest to answer C.

 


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