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Naily [24]
2 years ago
15

A gas with a pressure of 820.4 mmHg occupies a

Chemistry
1 answer:
irga5000 [103]2 years ago
5 0

Answer:

V₂ = 1223.2 mL  

Explanation:

Given data:

Pressure of gas = 820.4 mmHg

Initial volume of gas = 900.0 mL

Initial temperature = 25.0°C (25+273=298K)

Final temperature = 132.0°C (132.0 +273 = 405 K)

Final volume = ?

Solution:

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 900.0 mL × 405 K / 298 k

V₂ = 364500 mL.K / 298 K

V₂ = 1223.2 mL  

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In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very
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<u>Answer:</u> The final concentration of potassium nitrate is 5.70\times 10^{-6}M

<u>Explanation:</u>

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M

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

M_1V_1=M_2V_2          .......(1)

  • <u>Calculating for first dilution:</u>

M_1\text{ and }V_1 are the molarity and volume of the concentrated KNO_3 solution

M_2\text{ and }V_2 are the molarity and volume of diluted KNO_3 solution

We are given:

M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL

Putting values in equation 1, we get:

7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M

  • <u>Calculating for second dilution:</u>

M_2\text{ and }V_2 are the molarity and volume of the concentrated KNO_3 solution

M_3\text{ and }V_3 are the molarity and volume of diluted KNO_3 solution

We are given:

M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL

Putting values in equation 1, we get:

1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M

Hence, the final concentration of potassium nitrate is 5.70\times 10^{-6}M

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