What is the mole fraction of solute in a 3.89 m aqueous solution?

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What is the mole fraction of solute in a 3.89 m aqueous solution?

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Answer 1 · 2016-12-13T06:50:26

$Molality$ $"Molality"$ $=$ $=$ $\frac{Moles of solute}{Kilograms of solvent}$ $"Moles of solute"/"Kilograms of solvent"$

$χ = \frac{3.89 \cdot m o l}{3.89 \cdot m o l + \frac{1000 \cdot g}{18.01 \cdot g \cdot m o l^{- 1}}} = 0.067$ $chi=(3.89*mol)/(3.89*mol+(1000*g)/(18.01*g*mol^-1))=0.067$

Explanation:

And $χ,$ $chi,$ $mole fraction$ $"mole fraction"$ $=$ $=$ $\frac{Moles of solute}{Moles of solute and moles of solvent}$ $"Moles of solute"/"Moles of solute and moles of solvent"$

We have a $3.89$ $3.89$ $molal$ $"molal"$ solution (at least I think we do); i.e. $molality$ $"molality"$ $=$ $=$ $3.89 \cdot m o l \cdot k g^{- 1}$ $3.89*mol*kg^-1$ .

And thus $χ = \frac{3.89 \cdot m o l}{3.89 \cdot m o l + \frac{1000 \cdot g}{18.01 \cdot g \cdot m o l^{- 1}}} = 0.067$ $chi=(3.89*mol)/(3.89*mol+(1000*g)/(18.01*g*mol^-1))=0.067$

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What is the mole fraction of solute in a 3.89 m aqueous solution?

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