Problem 66 - General Chemistry: Principles and Modern Applications

Question

Without doing detailed calculations, determine which of the following samples occupies the largest volume: (a) 25.5 mol of sodium metal \(\left(d=0.971 \mathrm{g} / \mathrm{cm}^{3}\right)\) (b) 0.725 L of liquid bromine \((d=3.12 \mathrm{g} / \mathrm{mL})\) (c) \(1.25 \times 10^{25}\) atoms of chromium metal \(\left(d=9.4 \mathrm{g} / \mathrm{cm}^{3}\right)\) (d) \(2.15 \mathrm{kg}\) of plumber's solder \(\left(d=9.4 \mathrm{g} / \mathrm{cm}^{3}\right), \mathrm{a}\) lead-tin alloy with a 2: 1 atom ratio of lead to tin

Step-by-Step Solution

Answer

The sample that occupies the largest volume is option (c): 1.25 x 10^25 atoms of chromium metal

Step 1: Determine the volume for option (a)

Calculate the mass using the formula: \(m = n \times M_A\). Here \(n = 25.5 mol\) and \(M_A = 23 g/mol\) (for sodium). The calculated mass is \(m = 25.5 \times 23 = 586.5 g\). Now, calculate the volume: \(V = \frac{m}{d} = \frac{586.5}{0.971} = 604 \mathrm{cm}^3\)

Step 2: Determine the volume for option (b)

For liquid bromine, the volume is given directly as 0.725 L. But convert this to cm^3 as \(1 L = 1000 cm^3\). Therefore, \(V = 0.725 \times 1000 = 725 cm^3\)

Step 3: Determine the volume for option (c)

First calculate mass using \(m = n \times M_A\). Here, since we are given the number of atoms, we use the conversion \(1 mol = 6.022 \ x^{23}\) atoms to convert to moles. \(n = 1.25 x^{25} / 6.022 x^{23} = 207.66 mol\). Now, for chromium \(M_A = 52 g/mol\). Hence, \(m = n \times M_A = 207.66 x 52 = 10800 g\). Now, determine the volume: \(V = m/d = 10800/9.4 = 1149 cm^3\)

Step 4: Determine the volume for option (d)

Here, mass is given as 2.15 kg which is converted to grams as 2150 g. Now, calculate the volume: \(V = m/d = 2150/9.4 = 228.7 cm^3\).

Key Concepts

Mole Concept

The mole is a standard unit in chemistry used to express the amount of a chemical substance. It's similar to terms like "dozen" for twelve items, but instead, a mole refers to Avogadro's number, which is approximately 6.022 x 10^{23}.
This number is incredibly large because atoms and molecules are extremely small.
Using the mole concept, chemists can count atoms, molecules, and ions by weighing them. Let's take a quick look at how moles apply in determining volume. Consider option (a) where we have 25.5 moles of sodium. Knowing the molar mass of sodium ( 23 g/mol), we can calculate the mass using:

• Mass ( m) = Number of Moles ( n) x Molar Mass ( M_A)
• m = 25.5 x 23 = 586.5 g

This conversion from moles to grams allows us to work with density to find volume. Practicing this conversion helps you in balancing chemical equations and understanding stoichiometry.

Density in Chemistry

Density is a crucial physical property that relates the mass of a substance to its volume. In a chemical context, density is often used to calculate how much space an amount of material will occupy. The formula for density is:

• \(d = \frac{m}{V}\) , where \( m \) is mass and \( V \) is volume.

From this formula, once we have the mass (through prior mole and atoms conversion), we can isolate volume as:

• \(V = \frac{m}{d}\)

For instance, in option (c) with chromium, after determining mass by transferring from atom count to moles and subsequently grams, we apply the density variable to find volume. With a density of 9.4 g/cm^3, we divide the mass to identify \(V\). Each step emphasizes the density's use in predicting how much space substances occupy, aiding in practical applications such as storage and transferring of chemicals.

Chemical Element Properties

Chemical elements each have unique properties that affect how they react, their bonding, structure, and even common physical properties like density. When examining tasks like volume determination, understanding elemental properties, such as atomic weight and structure, is vital. For example, examining sodium versus bromine reveals that liquids, like bromine, often express volume differently compared to metals. Metals generally have crystalline structures affecting density, while liquids may spread out and occupy more space. Moreover in alloys, such as plumber's solder in option (d), the combination of elements (lead and tin) and their ratios determine the property's outcome, impacting the composite's density and response to environmental conditions like temperature. Recognizing these properties enhances our grasp on the predicted behavior of substances, for calculated dealings with compounds and elements in diverse environmental and experimental contexts.