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A student needs to prepare a 0.1 mg total mixture with a mole fraction of urea...

A student needs to prepare a 0.1 mg total mixture with a mole fraction of urea (Xurea) = 0.4673. How much TCA and urea (in mg) is needed for this mixture?

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Answer #1

Tatal mixture = 0.1 mg

Urea mole fraction = 0.4673

TCA mole fraction = 1 - 0.4673 = 0.5327

mole fraction of urea = mg of urea/(mg of urea +mg of TCA)

mg of urea = mole fraction of urea (mg of urea +mg of TCA)

= 0.4673 0.1 = 0.04673 mg

mg of urea = 0.04673 mg

mole fraction of TCA = mg of TCA/(mg of urea +mg of TCA)

mg of urea = mole fraction of TCA (mg of urea +mg of TCA)

= 0.5327 0.1 = 0.05327 mg

mg of TCA = 0.05327 mg

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