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Chapter 8: Problem 16

A quantity of \(1 \mathrm{~g}\) of metallic carbonate \(\mathrm{XCO}_{3}\) is completely converted into a chloride \(\mathrm{XCl}_{2}\) weighing \(1.11 \mathrm{~g}\). The atomic mass of the element ' \(\mathrm{X}^{\prime}\) is (a) 10 (b) 20 (c) 30 (d) 40

Short Answer

Expert verified
Answer: The atomic mass of element X is 40.

Step by step solution

01

Determine the mass of element X in both XCO3 and XCl2 compounds

First, let's find the mass of element X in both the given compounds. For XCO\(_3\), the mass of X is the given mass (\(1g\)) minus the mass of carbon (\(12g/mol\)) and 3 times the mass of oxygen (\(3*16g/mol\)). For XCl\(_2\), the mass of X is the given mass (\(1.11g\)) minus the mass of 2 times the mass of chlorine (\(2*35.5g/mol\)).
02

Write the ratio of the mass of X in both compounds

Now, we can write the ratio of the mass of X in XCO3 and XCl2: \(\frac{mass\, of\, X\, in\, XCO_3}{mass\, of\, X\, in\, XCL_2}=\frac{\Bigl( atomic\, mass\, of\, X - (12+3*16) \Bigr)}{\Bigl( atomic\, mass\, of\, X - 2*35.5 \Bigr)} = \frac{1}{1.11}\)
03

Simplify and solve the equation for the atomic mass of X

Simplify the equation to solve for the atomic mass of X: \((atomic\, mass\, of\, X - 60)/(atomic\, mass\, of\, X - 71) = \frac{1}{1.11}\) After cross-multiplying, we get: \(1.11*(atomic\, mass\, of\, X - 60) = (atomic\, mass\, of\, X - 71)\) Expanding and simplifying further: \(1.11*atomic\, mass\, of\, X - 66.6 = atomic\, mass\, of\, X - 71\) Moving the terms with X to one side and constants to another side: \(0.11*atomic\, mass\, of\, X = 4.4\) Now, solving for the atomic mass of X: \(atomic\, mass\, of\, X = \frac{4.4}{0.11} = 40\) Thus, the atomic mass of element X is \(\boxed{40}\), which corresponds to answer choice (d).

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Most popular questions from this chapter

A compound contains elements \(X\) and \(\mathrm{Y}\) in \(1: 4\) mass ratio. If the atomic masses of \(X\) and \(Y\) are in \(1: 2\) ratio, the empirical formula of compound should be (a) \(\mathrm{XY}_{2}\) (b) \(\mathrm{X}_{2} \mathrm{Y}\) (c) \(\mathrm{XY}_{4}\) (d) \(\mathrm{X}_{4} \mathrm{Y}\)

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