Question

The molefraction of water in mixture containing 9.0g of water and 120g of CH3COOH

Answer 1

Moles of water = 0.5

Moles of acid = 2

Hence mole fraction of water = 0.5/2.5

= 0.2.

Answer 2

Answer: The mole fraction of water in mixture is $0.2$.

Explanation:

Suppose in a mixture two components are present $1$ and $2$, therefore the mole fraction of component $1$ $(X_{1} )$ is:

$X_{1} = \frac{n_{1} }{n_{1}+ n_{2} }$

where $n_{1}$ = number of moles of component $1$

$n_{2}$ =  number of moles of component $2$.

Number of moles can be calculated using:

$n = \frac{given mass}{molecular mass}$

Now given mixture is water and acetic acid:

$n _{water } = \frac{9 g}{18 g mol^{-1} } = 0.5 moles$        ∵ Molecular mass of water = $18 g mol^{-1}$

$n _{acetic acid } = \frac{120 g}{60 g mol^{-1} } = 2 moles$  ∵ Molecular mass of acetic acid =$60 g mol^{-1}$

Mole fraction of water can now be calculated as:

$X_{water} = \frac{n_{water} }{n_{water}+ n_{acetic acid} }$

$X_{water} = \frac{0.5 moles }{0.5 moles + 2 moles } = 0.2$

Thus the mole fraction of water in mixture is $0.2$.

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