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3 years ago

Calculate the concentration of a solution formed by dissolving 5.6 g of potassium

Chemistry

1 answer:

gayaneshka [121]3 years ago

7 0

The concentration of solution : 0.2 M

<h3> Further explanation </h3>

Given

5.6 g Potassium hydroxide(KOH)

500 ml of solution = 0.5 L

Required

The concentration

Solution

Molarity shows the number of moles of solute in every 1 liter of solution or mmol in each ml of solution

MW KOH = 39+16+1=56 g/mol

mol solute(KOH) :

= mass : MW KOH

= 5.6 : 56 g/mol

= 0.1

Molarity :

= 0.1 : 0.5

= 0.2 M

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2 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:

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

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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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