(1) Calculate the weight of ethanol

$\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{etOH}\mathbf{}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\overline{)\mathbf{mL}}\mathbf{)}\mathbf{\left(}\frac{\mathbf{1}\mathbf{}\overline{){\mathbf{cm}}^{\mathbf{3}}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{789}\mathbf{}\overline{)\mathbf{g}}}{\overline{){\mathbf{cm}}^{\mathbf{3}}}}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{1}\mathbf{x}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{kg}}{\mathbf{1}\mathbf{}\overline{)\mathbf{g}}}\mathbf{\right)}$

**mass etOH = 1.1835x10 ^{-3} kg**

A mysterious white powder could be powdered sugar (C_{12}H_{22}O_{11}), cocaine (C_{17}H_{21}NO_{4}), codeine (C_{18}H_{21}NO_{3}), norfenefrine (C_{8}H_{11}NO_{2}), or fructose (C_{6}H_{12}O_{6}). When 82 mg of the powder is dissolved in 1.50 mL of ethanol (d = 0.789 g/cm^{3}, normal freezing point −114.6 ^{o}C, K_{f} = 1.99 ^{o}C/m), the freezing point is lowered to −115.5 ^{o}C. What is the identity of the white powder?

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Our tutors have indicated that to solve this problem you will need to apply the Freezing Point Depression concept. If you need more Freezing Point Depression practice, you can also practice Freezing Point Depression practice problems.

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Based on our data, we think this problem is relevant for Professor Blaha's class at Columbus State Community College.