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Chapter 1: Problem 66

Lithium is the least dense metal known (density: \(0.53 \mathrm{~g} / \mathrm{cm}^{3}\) ). What is the volume occupied by \(1.20 \times 10^{3} \mathrm{~g}\) of lithium?

Short Answer

Expert verified
The volume occupied by \(1.20 \times 10^{3} \mathrm{~g}\) of Lithium is \(2260 \mathrm{~cm^{3}}\).

Step by step solution

01

Understand the Problem

We are given that the density of lithium is \(0.53 \mathrm{~g/cm^{3}}\) and the mass of lithium is \(1.20 \times 10^{3} \mathrm{~g}\). We have to find the volume that this mass of lithium will occupy. The formula that connects these quantities is the formula for density, which is \(density = \frac{mass}{volume}\). To find volume, we'll rearrange the formula to isolate the term for volume.
02

Rearrange the Density Equation

Start off by rewriting the density formula and rearrange it to solve for volume. The rearranged formula will be \( volume = \frac{mass}{density} \).
03

Substitute Given Values

Next, we substitute the values of mass and density into our equation from step 2. The mass is \(1.20 \times 10^{3} \mathrm{~g}\) and the density is \(0.53 \mathrm{~g/cm^{3}}\). So, the equation becomes \( volume = \frac{1.20 \times 10^{3} \mathrm{~g}}{0.53 \mathrm{~g/cm^{3}}} \).
04

Solve the Equation

Upon carrying out the division in step 3, we find that the volume is approximately \(2264.15 \mathrm{~cm^{3}}\).
05

Round the Answer

The final answer should be in three significant figures, as our inputs are in three significant figures. So, after rounding, the final volume of lithium would be \(2260 \mathrm{~cm^{3}}\).

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