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VARVARA [1.3K]
3 years ago
6

A 1.30M solution of BaCl2 has a density of 1.230 g/ml. a) What is the mole fraction of BaCl2 in this solution?

Chemistry
1 answer:
Elodia [21]3 years ago
6 0

<u>Answer:</u> The mole fraction of barium chloride in the solution is 0.024

<u>Explanation:</u>

We are given:

Molarity of barium chloride solution = 1.30 M

This means that 1.30 moles of barium chloride is present in 1 L or 1000 mL of solution.

  • To calculate the mass of solution, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.230 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

1.230g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.230g/mL\times 1000mL)=1230g

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Moles of barium chloride = 1.30 moles

Molar mass of barium chloride = 208 g/mol

Putting values in equation 1, we get:

1.30mol=\frac{\text{Mass of barium chloride}}{208g/mol}\\\\\text{Mass of barium chloride}=(1.30mol\times 208g/mol)=270.4g

Mass of water = Mass of solution - Mass of barium chloride

Mass of water = 1230 - 270.4 = 959.6 g

<u>Calculating the moles of water:</u>

Given mass of water = 959.6 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

\text{Moles of water}=\frac{959.6g}{18g/mol}\\\\\text{Moles of water}=53.31mol

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

  • <u>For barium chloride:</u>

\chi_{\text{(barium chloride)}}=\frac{n_{\text{(barium chloride)}}}{n_{\text{(water)}}+n_{\text{(barium chloride)}}}

\chi_{\text{(barium chloride)}}=\frac{1.30}{1.30+53.31}\\\\\chi_{\text{(barium chloride)}}=0.024

Hence, the mole fraction of barium chloride in the solution is 0.024

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We convert mass to moles:

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if the Celsius temperature of a gas at constant pressure is increased from 10 Celsius to 20 Celsius the volume is
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In statement the data given is,

T₁  =  10 °C  =  283.15 K                    ∴  K  =  273.15 + °C

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So, it is clear that the temperature is being increased hence, we will find an increase in volume. Let us assume that the starting volume is 100 L, so,

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Now, we will arrange equation 1 for V₂ as,

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Hence, it is proved that by increasing temperature from 10 °C to 20 °C resulted in the increase of Volume from 100 L to 103.52 L.

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