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ryzh [129]
3 years ago
8

A gaseous mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a

) 20 b) 9.25 c) 25 d) 6.75
Chemistry
1 answer:
Semenov [28]3 years ago
3 0

<u>Answer:</u> The correct answer is Option d.

<u>Explanation:</u>

We are given:

Mass percentage of CH_4 = 20 %

So, mole fraction of CH_4 = 0.2

Mass percentage of C_2H_4 = 30 %

So, mole fraction of C_2H_4 = 0.3

Mass percentage of C_2H_2 = 35 %

So, mole fraction of C_2H_2 = 0.35

Mass percentage of C_2H_2O = 15 %

So, mole fraction of C_2H_2O = 0.15

We know that:

Molar mass of CH_4 = 16 g/mol

Molar mass of C_2H_4 = 28 g/mol

Molar mass of C_2H_2 = 26 g/mol

Molar mass of C_2H_2O = 48 g/mol

To calculate the average molecular mass of the mixture, we use the equation:

\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}

where,

\chi_i = mole fractions of i-th species

m_i = molar masses of i-th species

n_i = number of observations

Putting values in above equation:

\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}

\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75

Hence, the correct answer is Option d.

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According to Huckel 's rule ,

The cyclic compounds with ( 4n+2 ) π electrons , where , n = 0, 1, 2, 3 ... , are considered to be aromatic in nature ,

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Now, for the cyclopropene structure , the number of π electron are equal to 4 , 2 from the double bond and 2 from the negative charge .

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a 2.7 L of N2 is collected at 121kpa and 288 K . if the pressure increases to 202 kpa and the temperature rises to 303 K , what
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Answer:

The gas will occupy a volume of 1.702 liters.

Explanation:

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P\cdot V = n\cdot R_{u}\cdot T (1)

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P - Pressure, measured in kilopascals.

V - Volume, measured in liters.

n - Molar quantity, measured in moles.

T - Temperature, measured in Kelvin.

R_{u} - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.

We can simplify the equation by constructing the following relationship:

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Where:

P_{1}, P_{2} - Initial and final pressure, measured in kilopascals.

V_{1}, V_{2} - Initial and final volume, measured in liters.

T_{1}, T_{2} - Initial and final temperature, measured in Kelvin.

If we know that P_{1} = 121\,kPa, P_{2} = 202\,kPa, V_{1} = 2.7\,L, T_{1} = 288\,K and T_{2} = 303\,K, the final volume of the gas is:

V_{2} = \left(\frac{T_{2}}{T_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot V_{1}

V_{2} = 1.702\,L

The gas will occupy a volume of 1.702 liters.

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