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4 years ago

10

When a positive ion is formed, the resulting ion is

1 answer:

avanturin [10]4 years ago

3 0

A positive ion is also known as a cation, if that’s what you’re looking for.

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3 years ago

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?

Elodia [21]

<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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4 years ago

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