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Chapter 3: Problem 129

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 \(\mathrm{XeF}_{\text {a }}\) weigh \(0.368 \mathrm{~g}\), determine the value for \(n\) in the formula.

Short Answer

Expert verified
In the given xenon fluoride compound, the value of "n" in the formula \(\mathrm{XeF}_{n}\) is \(n = 2\), making the compound \(\mathrm{XeF}_{2}\). This was determined by finding the molecular weight of the compound using the given information and applying Avogadro's constant.

Step by step solution

01

Recall the formula relating weight, moles, and molecular weight

In chemistry, the relationship between the weight of a substance, the number of moles, and its molecular weight can be described using the formula: Weight = (Number of moles) × (Molecular weight) We will be using this formula to determine the molecular weight of the given xenon fluoride compound.
02

Convert the given weight into moles

We know the weight of \(9.03 \times 10^{20}\) molecules of XeF_a is 0.368 grams. To determine the number of moles, we will divide the weight by the molecular weight. The molecular weight of XeF_n can be given as: Molecular weight of XeF_n = (Xe's Molecular weight) + n × (F's Molecular weight) Where weights of Xe and F are known molecular weights: Xe = 131.29 g/mol and F = 18.998 g/mol. Number of moles = \(\frac{0.368 g}{(131.29 g/mol + n \times 18.998 g/mol)}\)
03

Find the number of moles of XeF_n using Avogadro's constant

After calculating the moles of XeF_n, we can determine the number of moles present in the given compound by using Avogadro's constant. The constant is: Avogadro's constant = \(6.022 \times 10^{23}\) molecules/mol Number of moles = \(\frac{9.03 \times 10^{20} \text{ molecules}}{6.022 \times 10^{23} \text{ molecules/mol}}\) Number of moles ≈ \(1.5 \times 10^{-3} \text{ mol}\)
04

Find the molecular weight and the value of n in the XeF_n formula

Now that we have the number of moles of XeF_n, we can substitute it back into the original equation of: Weight = (Number of moles) × (Molecular weight) To solve for n: \(0.368 g = (1.5 \times 10^{-3} \text{ mol}) \times (131.29 g/mol + n \times 18.998 g/mol)\) Solving for n, we get: \(n \approx 1.98\) Since n must be a whole number, we can round up to the nearest whole number: \(n = 2\) Therefore, the value of "n" in the formula for the given xenon fluoride compound is \(n = 2\). The compound is \(\mathrm{XeF}_{2}\).

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Most popular questions from this chapter

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