Question

1.3 chemical problem solving
what is the density of a gold coin if it has a mass of 6.2 g, a diameter of 2.51 cm, and a height of 0.050 cm?
○ 20 g/cm³
○ 36.27 g/cm³
○ 16 g/cm³
○ 25 g/cm³

Explanation:

Step1: Recall the formula for density and the volume of a cylinder

Density ($
ho$) is given by the formula $
ho = \frac{m}{V}$, where $m$ is mass and $V$ is volume. The gold coin is a cylinder, so the volume of a cylinder is $V=\pi r^{2}h$, where $r$ is the radius and $h$ is the height. The radius $r$ is half of the diameter, so $r=\frac{d}{2}$.

Step2: Calculate the radius of the gold coin

Given the diameter $d = 2.51\space cm$, the radius $r=\frac{2.51}{2}=1.255\space cm$.

Step3: Calculate the volume of the gold coin

Using the volume formula for a cylinder $V = \pi r^{2}h$. Substitute $r = 1.255\space cm$, $h=0.050\space cm$ and $\pi\approx3.14$.
\[

$$\begin{align*} V&=3.14\times(1.255)^{2}\times0.050\\ &=3.14\times1.575025\times0.050\\ &=3.14\times0.07875125\\ &\approx0.247\space cm^{3} \end{align*}$$

\]

Step4: Calculate the density of the gold coin

Using the density formula $
ho=\frac{m}{V}$, where $m = 6.2\space g$ and $V\approx0.247\space cm^{3}$.
\[

ho=\frac{6.2}{0.247}\approx25\space g/cm^{3}
\]

Answer:

25 g/cm³ (The option is: 25 g/cm³)