Q. 2.120 - An Introduction to General, Organic, and Biological Chemistry

Question

A $50.0$-g silver object and a $50.0$-g gold object are both added to $75.5 \mathrm{mL}$ of water contained in a graduated cylinder. What is the new water level, in milliliters, in the cylinder (see Table 2.8) ? (2.7)

Step-by-Step Solution

Answer

The new water level in millimeters in the cylinder is determined as $82.9 mL$.

Step 1 : Given Information

The given is the graduated cylinder filled with silver and golden objects.

The objective is to determine the new level of water in millimeters in cylinder.

We are given the following:

The mass of a silver object is 50 g. 

The mass of gold object is 50 g. 

The volume of water is  75.5 mL. 

Step 2 : Find the values from the table

We know that

Density of silver $=10.5 \mathrm{g}/\mathrm{mL}$

Density of gold $=19.3 \mathrm{g}/\mathrm{mL}$

Step 3 : Explanation

The formula for density can be used to compute the volume of water, silver, and gold.

The formula for density is as follows:

$Density =\frac{ mass of object}{volume of object }$

By rearranging the above equation, the volume of the item may be computed.

$Volume of object =\frac{mass of object}{Density of object }$

As a result of the aforesaid values, we get

$$\begin{gathered} Volume of silver=\frac{50.0 g}{10.5 g/ mL} \\ =4.76 mL \end{gathered}$$

$$\begin{gathered} Volume of gold=\frac{50.0 g}{19.3 g /mL} \\ =2.59 mL \end{gathered}$$

Step 4 : Explanation

As a result, the total amount of silver and gold is as follows:

$Volume for silver and gold =4.76 mL+2.59 mL$

$=7.35 \mathrm{mL}$

As a result, the amount of water displaced by a silver and gold coin in water is $7.35 \mathrm{mL}$.

$Total volume of water = water level before object submerged + water displaced$

$$\begin{gathered} =75.5 mL+7.35 mL \\ =82.9 mL \end{gathered}$$

So, the new water level in the cylinder as $82.9 \mathrm{mL}$.