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1 year ago

11

Suppose you measure the absorbance of a yellow dye solution in a 1.00 cm cuvette. The absorbance of the solution at 427 nm is 0.

20. If the molar absorptivity of yellow dye at 427 nm is 27400 M–1cm–1, what is the concentration of the solution?

1 answer:

Dimas [21]1 year ago

4 0

Answer:

The concentration of the solution, $C=7.2992\times 10^{-6} M$

Explanation:

The absorbance of a solution can be calculated by Beer-Lambert's law as:

$A=\varepsilon Cl$

Where,

A is the absorbance of the solution

ɛ is the molar absorption coefficient ( $L.mol^{-1}.cm^{-1}$ )

C is the concentration ( $mol^{-1}.L^{-1}$ )

l is the path length of the cell in which sample is taken (cm)

Given,

A = 0.20

ɛ = 27400 $M^{-1}.cm^{-1}$

l = 1 cm

Applying in the above formula for the calculation of concentration as:

$A=\varepsilon Cl$

$0.20= 27400\times C\times 1$

$C = \frac{0.20}{27400\times 1} M$

So , concentration is:

$C=7.2992\times 10^{-6} M$

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Draw the best lewis structure for bro4- and determine the formal charge on bromine

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Answer :

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Answer:

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Answer:

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2 years ago

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes.a. 0.0200 mol sodium p

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<u>Answer:</u>

<u>For a:</u> The concentration of $Na^+\text{ and }PO_4^{3-}$ ions in the solution are 6 M and 2 M respectively.

<u>For b:</u> The concentration of $Ba^{2+}\text{ and }NO_3^{-}$ ions in the solution are 0.5 M and 1.0 M respectively.

<u>For c:</u> The concentration of $K^{+}\text{ and }Cl^{-}$ ions in the solution are 0.051 M and 0.051 M respectively.

<u>For d:</u> The concentration of $NH_4^{+}\text{ and }SO_4^{2-}$ ions in the solution are 1.34 M and 0.67 M respectively.

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ ......(1)

Or,

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$ ...(2)

For the given options:

<u>For a:</u>

The chemical formula of sodium phosphate is $Na_3PO_4$

Moles of sodium phosphate = 0.0200 moles

Volume of solution = 10.0 mL

Putting values in equation 1, we get:

$\text{Molarity of the sodium phosphate}=\frac{0.0200\times 1000}{10.0}=2M$

1 mole of sodium phosphate produces 3 moles of $Na^+$ ions and 1 mole of $PO_4^{3-}$ ions

So, concentration of $Na^+\text{ ions}=(3\times 2)=6M$

Concentration of $PO_4^{3-}\text{ ions}=(1\times 2)=2M$

Hence, the concentration of $Na^+\text{ and }PO_4^{3-}$ ions in the solution are 6 M and 2 M respectively.

<u>For b:</u>

The chemical formula of barium nitrate is $Ba(NO_3)_2$

Moles of barium nitrate = 0.300 moles

Volume of solution = 600.0 mL

Putting values in equation 1, we get:

$\text{Molarity of the barium nitrate}=\frac{0.300\times 1000}{600.0}=0.5M$

1 mole of barium nitrate produces 1 mole of $Ba^{2+}$ ions and 2 mole of $NO_3^{-}$ ions

So, concentration of $Ba^{2+}\text{ ions}=(1\times 0.5)=0.5M$

Concentration of $NO_3^{-}\text{ ions}=(2\times 0.5)=1M$

Hence, the concentration of $Ba^{2+}\text{ and }NO_3^{-}$ ions in the solution are 0.5 M and 1.0 M respectively.

<u>For c:</u>

The chemical formula of potassium chloride is KCl

Given mass of potassium chloride = 1.00 g

Molar mass of potassium chloride = 39 g/mol

Volume of solution = 0.500 L

Putting values in equation 1, we get:

$\text{Molarity of the potassium chloride}=\frac{1.00}{39\times 0.500}=0.051M$

1 mole of potassium chloride produces 1 mole of $K^{+}$ ions and 1 mole of $Cl^{-}$ ions

So, concentration of $K^{+}\text{ ions}=(1\times 0.051)=0.051M$

Concentration of $Cl^{-}\text{ ions}=(1\times 0.051)=0.051M$

Hence, the concentration of $K^{+}\text{ and }Cl^{-}$ ions in the solution are 0.051 M and 0.051 M respectively.

<u>For d:</u>

The chemical formula of ammonium sulfate is $(NH_4)_2SO_4$

Given mass of ammonium sulfate = 132 g

Molar mass of ammonium sulfate = 132 g/mol

Volume of solution = 1.50 L

Putting values in equation 1, we get:

$\text{Molarity of the ammonium sulfate}=\frac{132}{132\times 1.50}=0.67M$

1 mole of ammonium sulfate produces 2 moles of $NH_4^{+}$ ions and 1 mole of $SO_4^{2-}$ ions

So, concentration of $NH_4^{+}\text{ ions}=(2\times 0.67)=1.34M$

Concentration of $SO_4^{2-}\text{ ions}=(1\times 0.67)=0.67M$

Hence, the concentration of $NH_4^{+}\text{ and }SO_4^{2-}$ ions in the solution are 1.34 M and 0.67 M respectively.

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