Q.1 - An Introduction to General, Organic, and Biological Chemistry

Question

Gold, one of the most sought-after metals in the world, has a density of $19.3\mathrm{g}/{\mathrm{cm}}^3$, a melting point of ${1064}^{\circ }\mathrm{C}$, and specific heat of $0.129\mathrm{J}/{\mathrm{g}}^{\circ }\mathrm{C}$. A gold nugget found in Alaska in 1998 weighed 20.17 lb.

a. How many significant figures are in the measurement of the weight of the nugget?

b. Which is the mass of the nugget in kilograms?

c. If the nugget were pure gold, what would its volume be in cubic centimetres?

d. What is the melting point of gold in degrees Fahrenheit and kelvins?

e. How many kilocalories are required to raise the temperature of the nugget from ${500}^{\circ }\mathrm{C}\text{ to }{1064}^{\circ }\mathrm{C}$ ?

f. If the price of gold is $\mathrm{\$}42.06$ per gram, what is the nugget worth in dollars.

Step-by-Step Solution

Answer

a. $4$ significant figures are in the measurement of the weight of the nugget 

b. $9.17kg$

c. $475c{m}^3$

d. ${1947}^{\circ }\mathrm{F} \mathrm{and} 1337\mathrm{K}$

e. $159kcal$

f. $\$386000$

Step 1: Given Information

Given,

The density of gold = $19.3\mathrm{g}/{\mathrm{cm}}^3$

Specific heat = $0.129\mathrm{J}/{\mathrm{g}}^{\circ }\mathrm{C}$

Melting point = ${1064}^{\circ }\mathrm{C}$

Step 2: Explanation Part (a)

The weight of the nugget is$20.17lb$

It has four significant figures in the measurement of the weight of the nugget because zero and non-zero digits present between non zero digits are significant.

Step 3: Explanation Part (b)

The weight of the nugget is $20.17lb$

now, the weight in kg is,

$$\begin{gathered} 1\mathrm{kg}=2.20\mathrm{lb} \\ \Rightarrow 20.17 \mathrm{lb}=\frac{20.17}{2.20} \\ =9.17\mathrm{kg} \end{gathered}$$

Step 4: Explanation Part (c)

We know,

The density of gold is$19.3\mathrm{g}/{\mathrm{cm}}^3$

and mass is $9.17kg=9170g$

Volume = $\frac{mass}{density}$

$\Rightarrow \frac{1970}{19.3}=475c{m}^3$

Step 5: Explanation Part (d)

We know,

the melting point of gold = ${1064}^{\circ }\mathrm{C}$

the melting point of gold in Fahrenheit is$T_{\mathrm{F}}=1.8\left(T_{\mathrm{C}}\right)+32$

$=1.8\left(1064\right)+32={1947}^{o}F$

the melting point of gold in kelvin is $T\left(\mathrm{K}\right)=T\left({}^{\circ }\mathrm{C}\right)+273$

$=1064+273={1337}^{o}K$

Step 6: Explanation Part (e)

We know,

Mass = $9.17kg$

specific heat $c=0.129\mathrm{J}/{\mathrm{g}}^{\circ }\mathrm{C}$

Change in temperature$∆T={500}^{\circ }\mathrm{C}\text{ to }{1064}^{\circ }\mathrm{C}={564}^{\mathrm{o}}\mathrm{C}$

$1\mathrm{cal}=4.184\mathrm{J}$

The calories required are,

$Q=mc\mathrm{\Delta }T$

$\frac{\Rightarrow 9170\times 564\times 0.129}{4.184\times 1000}=159kcal$

Step 7: Explanation Part (f)

We know,

price of gold = $\mathrm{\$}42.06\text{ per gram }$

Price of nugget 

 $$\begin{gathered} =9170\times 42.06 \\ =\$386000 \end{gathered}$$