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Chapter 3: Problem 137

A given sample of a xenon fluoride compound contains molecules of the type \(\mathrm{XeF}_{n},\) where \(n\) is some whole number. Given that $9.03 \times 10^{20}\( molecules of \)\mathrm{XeF}_{n}\( weigh \)0.368 \mathrm{g},$ determine the value for \(n\) in the formula.

Short Answer

Expert verified
The value of \(n\) in the formula of the xenon fluoride compound, XeF_n, is 6, and the compound is XeF_6.

Step by step solution

01

Find the number of moles of the xenon fluoride compound

To do this, we'll use Avogadro's number (approximately \(6.022 \times 10^{23}\) molecules per mole) and the given number of molecules: \(9.03 \times 10^{20} \, \text{molecules}\). Divide the given number of molecules by Avogadro's number to get the number of moles: \[ \frac{9.03 \times 10^{20} \, \text{molecules}}{6.022 \times 10^{23} \, \text{molecules/mol}} = 1.5 \times 10^{-3} \, \text{mol} \]
02

Determine the molar mass of the xenon fluoride compound

Divide the given mass of the compound (0.368 g) by the number of moles (calculated in step 1): \[ \frac{0.368 \, \text{g}}{1.5 \times 10^{-3} \, \text{mol}} = 245.33 \, \text{g/mol} \]
03

Find the molar mass of xenon (Xe) and fluoride (F)

The molar mass of xenon is approximately 131.29 g/mol, and the molar mass of fluoride is approximately 19.00 g/mol.
04

Determine the value of n in the formula XeF_n

By knowing the molar mass of the xenon fluoride compound, the molar mass of xenon, and the molar mass of fluoride, we can write the following equation for the molar mass of XeF_n: \[ 131.29 + 19.00n = 245.33 \] Now, we can solve for the value of n: \[ n = \frac{245.33 - 131.29}{19.00} = 6 \] So, the value of n in the formula XeF_n is 6, and the compound is XeF_6.

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