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3 years ago

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A mixture of carbon dioxide and helium gases is maintained in a 7.91 L flask at a pressure of 1.42 atm and a temperature of 33 °

C. If the gas mixture contains 8.25 grams of carbon dioxide, the number of grams of helium in the mixture is g.

1 answer:

Crank3 years ago

8 0

Answer:

The gas mixture contains 1.038 grams of helium

Explanation:

<u>Step 1:</u> Data given

Volume of the flask = 7.91 L

Total Pressure = 1.42 atm

Temperature = 33 °C

Mass of CO2 = 8.25 grams

Molar mass of CO2 = 44.01 g/mol

Molar mass of He = 4 g/mol

<u>Step 2</u>: Calculate total number moles of gas

p*V = n*R*T

⇒ p = the pressure = 1.42 atm

⇒ V = the volume = 7.91 L

⇒ n= the number of moles = TO BE DETERMINED

⇒ R = the gas constant = 0.08206 L* atm/K*mol

⇒ T = the temperature = 33 °C = 306 Kelvin

n = (p*V)/(R*T)

n = (1.42*7.91)/(0.08206 * 306)

n = 0.447 moles

<u>Step 3</u>: Calculate moles of CO2

Moles CO2 = mass CO2 / Molar mass CO2

Moles CO2 = 8.25 grams / 44.01 g/mol

Moles CO2 = 0.1875 moles

<u>Step 4:</u> Calculate moles of Helium

Moles Helium = total moles of gas - moles of CO2

Moles Helium = 0.447 - 0.1875 = 0.2595 moles of helium

<u>Step 5: </u> Calculate mass of helium

Mass of helium = moles of helium * molar mass of helium

Mass of helium = 0.2595 moles * 4 g/mol

Mass of helium = 1.038 grams

The gas mixture contains 1.038 grams of helium

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A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

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This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

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Now we have to calculate the molar mass of unknown compound.

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Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

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If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

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Mass of O = 38.1 g

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Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

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For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

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Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

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