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Advocard [28]
2 years ago
7

How many grams of argon would it take to fill a large light bulb with a volume of 0.745 L at STP?

Chemistry
1 answer:
EastWind [94]2 years ago
8 0

Answer:

Mass = 1.33 g

Explanation:

Given data:

Mass of argon required = ?

Volume of bulb = 0.745 L

Temperature and pressure = standard

Solution:

We will calculate the number of moles of argon first.

Formula:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

By putting values,

1 atm ×0.745 L = n × 0.0821 atm.L/mol.K× 273.15 K

0.745 atm. L = n × 22.43 atm.L/mol

n = 0.745 atm. L / 22.43 atm.L/mol

n = 0.0332 mol

Mass of argon:

Mass = number of moles × molar mass

Mass = 0.0332 mol × 39.95 g/mol

Mass = 1.33 g

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To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}       ........(1)

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We are given:

0.01 M NaCl solution:

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C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1

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\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M

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Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{  (known)}\\\mu=0.01M

Putting values in equation 1, we get:

-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676

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We are given:

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C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1

Putting values in equation 2, we get:

\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M

Now, calculating the activity coefficient of K^{+} ion in the solution by using equation 1:

Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{  (known)}\\\mu=0.025M

Putting values in equation 1, we get:

-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851

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Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{  (known)}\\\mu=0.06M

Putting values in equation 1, we get:

-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794

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rearrange above equation

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