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Chapter 8: Problem 53

\(\begin{array}{l}\text { A gaseous mixture contains } 70 \% & \mathrm{~N}_{2}\end{array}\) and \(30 \%\) unknown gas, by volume. If the average molecular mass of gaseous mixture is \(37.60\), the molecular mass of unknown gas is (a) \(42.2\) (b) 60 (c) 40 (d) 50

Short Answer

Expert verified
Answer: The molecular mass of the unknown gas is 60 g/mol.

Step by step solution

01

Percentage Contribution

Using the given percentages, we know Nitrogen gas (\(N_2\)) contributes 70% and the unknown gas contributes 30% to the total mixture. We'll use these percentages to calculate their individual contributions to the mixture's molecular mass.
02

Molecular Mass of Nitrogen

The molecular mass of Nitrogen gas (\(N_2\)) is 28 g/mol (\(2\times14\) given Nitrogen atom has a mass of 14 g/mol). So, the contribution of Nitrogen gas to the average molecular mass of the gaseous mixture is \(0.7\times28=19.6\) g/mol.
03

Molecular Mass of Unknown Gas

Let's denote the molecular mass of the unknown gas as 'x'. We know that Nitrogen contributes 19.6 g/mol to the average molecular mass of the mixture and the unknown gas contributes 30% (0.3). Therefore, we can set up the equation: \(0.7\times28 + 0.3\times x=37.6\)
04

Solve for 'x'

Now, we just need to solve for 'x': \(0.3\times x=37.6 - 0.7\times28\) \(0.3\times x=37.6 - 19.6\) \(0.3\times x=18\) Now, we can solve for 'x': \(x=\frac{18}{0.3}=60\) So, the molecular mass of the unknown gas is 60 g/mol. The correct answer is (b) 60.

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