Question

When insufficient gastric acidity dilute hydrochloric acid is used with a mass fraction of HCl 8%.
Calculate the mole fractions of HCl and water in this solution.

Answer 1

Answer:

hope it helps you

Explanation:

To make the calculations easier, pick a sample of this solution that has a mass of exactly

100 g

. Since you already know that this sample contains

20 g

of hydrochloric acid, you can say that the mass of water, the solvent, will be

mass of solvent



mass H

2

O

=

mass of solution



100 g

−

mass of solute



20 g

mass H

2

O

=

80 g

Nest, use the molar mass of hydrochloric acid and the molar mass of water to convert the masses to moles.

20

g

⋅

1 mole HCl

36.46

g

=

0.5485 moles HCl

80

g

⋅

1 mole H

2

O

18.015

g

=

4.4407 moles H

2

O

Now, the mole fraction of hydrochloric acid is given by the ratio that exists between the number of moles of hydrochloric acid and the total number of moles present in the solution.

χ

HCl

=

moles HCl

total moles

In this case, the mole fraction of hydrochloric acid will be

χ

HCl

=

0.5485

moles

(

0.5485

+

4.4407

)

moles

=

0.1

−−−

To find the mole fraction of water, use the fact that the mole fractions of the solute and of the solute, respectively, must add up to give

1

.

χ

H

2

O

=

1

−

χ

HCl

You will find

χ

H

2

O

=

1

−

0.1

=

0.9

−−−

Answer 2

Answer:

The mole fractions of HCl, and H₂O in the solution is 0.04, and 0.96 respectively.

Explanation:

Given,

The mass fraction of HCl is 8%. i.e. if the solution is having 100 g then it contains 8g of HCl (solute), and 92g of H₂O (solvent).

To find,

The mole fraction of HCl and Water in the solution.

Calculation,

Number of moles of HCl present in the solution is:

$n_{HCl} = \frac{W_{HCl}}{G.M.W_{HCl}}$

$\implies n_{HCl} = \frac{8 g}{36.5 g}$

$n_{HCl}$ = 0.219

Number of moles of H₂O present in the solution is:

$n_{H_2O} = \frac{W_{H_2O}}{G.M.W_{H_2O}}$

$\implies n_{H_2O} = \frac{92 g}{18 g}$

$n_{H_2O}$ = 5.11

The mole fraction of HCl is:

$\chi_{HCl} = \frac{n_{HCl}}{n_{HCl} + n_{H_2O}}$

$\implies \chi_{HCl} = \frac{0.219}{0.219 + 5.11}$

⇒ $\chi_{HCl}$ = 0.04

Hence, $\chi_{H_2O} = 1 - \chi_{HCl}$

$\implies \chi_{H_2O} = 1 - 0.04$

$\implies \chi_{H_2O}$ = 0.96

Therefore, the mole fractions of HCl, and H₂O in the solution is 0.04, and 0.96 respectively.

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