A sample of metallic element X


A sample of metallic element X, weighing 4.315 g, combines with 0.4810 L of Cl_{2} gas (at normal pressure and 20.0^{o}C) to form the metal chloride with the formula XCl. If the density of Cl_2 gas under these conditions is 2.948 g/L, what is the mass of the chlorine? The atomic mass of chlorine is 35.453 amu. What is the atomic mass of X? What is the identity of X?

Solution:

We will begin by writing down what we know.

Weight of sample metallic element X = 4.315 g

Amount of Cl_2 gas = 0.4810 L

Density of gas = 2.948 g/L

Atomic mass of Cl = 35.453 amu

To find the mass of Cl, we will multiply the amount of Cl gas by the density.

Mass of Cl= 0.4810 L \times \frac{2.948 g}{1 L}

Mass of Cl= 1.41799 g

 

To find the atomic mass, we can multiply the atomic mass of the Cl by the ratio of the weight of the metallic element X to the mass of Cl as follows:

35.453 \text{amu Cl}\times \frac{4.315 g}{1.41799 g} =107.88 amu

 

Thus, we know that the atomic mass of X is 107.88 amu. Looking at the periodic table, we can see that the element X is in fact silver (Ag).

This question can be found in General Chemistry, 9th edition, chapter 2, question 2.148

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