Question

A mixture contains 23g of ethanol and 36g
of water some acetic acid is added to
the mixture and its mole fraction
becomes 0.5, how much acetic acid was added.​

Answer 1

A mixture contains 23g of ethanol and 36g

of water some acetic acid is added to

the mixture and its mole fraction

becomes 0.5, how much acetic acid was added.

Answer 2

Answer:

The amount of acetic acid equals to 150g was added in the mixture.

Explanation:

Given, the mass of ethanol in a mixture = 23g

The molar mass of the ethanol = 46g/mol

The number of moles of the ethanol $\big n_A=\frac{23}{46}=0.5$ moles

the mass of water = 36g

The molar mass of the water = 18g/mol

The number of moles of the water $\big n_B=\frac{36}{18}=2$ moles

Now, some amount of acetic acid added in the mixture.

Given, the mole fraction of acetic acid, $\big X_c$ = 0.5

Consider that 'x' is the number of moles of acetic acid.

The mole fraction of acetic acid is given by:

$\bid X_C=\frac{n_A}{n_A+n_B+n_C}$

$0.5=\frac{n_c}{n_c+0.5+2}$

$0.5\times n_c+0.5\times 2.5=n_c$

$n_c-0.5n_c=1.25$

$n_c=\frac{1.25}{0.5}$

$\big n_c=2.5\; moles$

Therefore, the moles of acetic acid added = 2.5 moles

We know that, the molar mass of acetic acid = 60g/mol

The mass of acetic acid = moles × molar mass = 2.5 × 60 = 150g

Therefore, 150g of acetic acid was added.

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