What are the mole fractions of the components of the solution formed when 92 g glycerol is mixed with 90 g water?

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What are the mole fractions of the components of the solution formed when 92 g glycerol is mixed with 90 g water?

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Answer 1 · 2014-06-25T17:15:01

You can read how to calculate mole fractions in:
How do you calculate mole fractions?

In your problem,

$n_{glycerol}$ $n_"glycerol"$ = 92 g glycerol × $\frac{1 mol glycerol}{92.09 g glycerol}$ $(1"mol glycerol")/(92.09"g glycerol")$ = 0.9990 mol glycerol (2 significant figures + 2 guard digits)

$n_{water}$ $n_"water"$ = 90 g water × $\frac{1 mol water}{18.02 g water}$ $(1"mol water")/(18.02"g water")$ = 4.994 mol water

$n_{total} = n_{glycerol} + n_{water}$ $n_"total" = n_"glycerol" + n_"water"$ = 0.9990 mol + 4.994 mol = 5.993 mol

The mole fraction $X$ $Χ$ of glycerol is

$X_{glycerol} = \frac{n_{glycerol}}{n_{total}} = \frac{0.9990 mol}{5.993 mol}$ $Χ_"glycerol" = n_"glycerol"/n_"total" = (0.9990"mol")/(5.993"mol")$ = 0.17

The mole fraction of water is

$X_{water} = \frac{n_{water}}{n_{total}} = \frac{4.994 mol}{5.993 mol}$ $Χ_"water" = n_"water"/n_"total" = (4.994"mol")/(5.993"mol")$ = 0.83

Note: The required answers can have only 2 significant figures, because that is all you gave for the masses of glycerol and water.

Hope this helps.

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What are the mole fractions of the components of the solution formed when 92 g glycerol is mixed with 90 g water?

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