Select the correct answer:

A 46.9 gram sample of a substance has a volume of about 3.5 centimeters [tex]$^3$[/tex]. It is solid at a room temperature of [tex]$23^{\circ} C$[/tex]. Out of the four substances whose properties are given, which is the most likely identity of this substance?

\begin{tabular}{|l|c|c|c|}
\hline \multicolumn{1}{|c|}{ Substance } & \begin{tabular}{c}
Density \\
[tex]$( g / cm^3 )$[/tex]
\end{tabular} & \begin{tabular}{c}
Melting \\
Point \\
[tex]$({ }^{\circ} C )$[/tex]
\end{tabular} & \begin{tabular}{c}
Boiling \\
Point \\
[tex]$({ }^{\circ} C )$[/tex]
\end{tabular} \\
\hline molybdenum & 10.28 & 2,623 & 4,639 \\
\hline mercury & 13.53 & -39 & 357 \\
\hline hafnium & 13.31 & 2,233 & 4,603 \\
\hline lead & 11.34 & 327 & 1,749 \\
\hline
\end{tabular}

A. molybdenum
B. mercury
C. hafnium
D. lead



Answer :

To identify the substance from the given options, we need to calculate the density of the substance using the given mass and volume and then compare the calculated density with the densities provided for each substance.

Here are the given values:
- Mass (weight) of the substance: 46.9 grams
- Volume of the substance: 3.5 cm³

First, calculate the density of the substance using the formula:
[tex]\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \][/tex]

Plugging in the given values:
[tex]\[ \text{Density} = \frac{46.9 \text{ grams}}{3.5 \text{ cm}^3} \][/tex]
[tex]\[ \text{Density} \approx 13.4 \text{ g/cm}^3 \][/tex]

Next, we'll compare this calculated density to the densities of the given substances:
- Molybdenum: 10.28 g/cm³
- Mercury: 13.53 g/cm³
- Hafnium: 13.31 g/cm³
- Lead: 11.34 g/cm³

The calculated density (13.4 g/cm³) is closest to the density of hafnium (13.31 g/cm³).

Therefore, based on the comparison of densities, the most likely identity of the substance is:
C. hafnium