Question

word problems

1. a block of aluminum occupies a volume of 15.0ml and weighs 40.5g. what is its density?
2. mercury metal is poured into a graduated cylinder that holds exactly 22.5ml. the mercury used to fill the cylinder weighs 306.0g. from this information, calculate the density of mercury.
3. what is the weight of the ethanol that exactly fills a 200.0ml container? the density of ethanol is 0.789g/ml.
4. a rectangular block of copper metal weighs 1896g. the dimensions of the block are 8.4cm by 5.5cm, by 4.6cm. from this data, what is the density of copper? (hint: find the volume of a block first)
5. what volume of silver metal will weigh exactly 2500.0g. the density of silver is 10.5g/cm³.
6. find the mass of 250.0ml of benzene. the density of benzene is 0.8765g/ml.
7. a block of lead has dimensions of 4.5cm by 5.2cm by 6.0cm. they block weighs 1587g. from this information, calculate the density of lead.
8. 28.5g of iron shot is added to a graduated cylinder containing 45.50ml of water. the water level rises to the 49.10ml mark. from this information, calculate the density of iron.

Explanation:

Step1: Recall density formula

The formula for density $
ho$ is $
ho=\frac{m}{V}$, where $m$ is mass and $V$ is volume.

Step2: Solve problem 1

Given $m = 40.5g$ and $V=15.0mL$, then $
ho=\frac{40.5g}{15.0mL}=2.7g/mL$.

Step3: Solve problem 2

Given $m = 306.0g$ and $V = 22.5mL$, then $
ho=\frac{306.0g}{22.5mL}=13.6g/mL$.

Step4: Solve problem 3

Given $
ho=0.789g/mL$ and $V = 200.0mL$, using $m=
ho V$, we have $m=0.789g/mL\times200.0mL = 157.8g$.

Step5: Solve problem 4

First find volume $V$ of the rectangular - block. $V=l\times w\times h=8.4cm\times5.5cm\times4.6cm = 212.52cm^{3}$. Given $m = 1896g$, then $
ho=\frac{1896g}{212.52cm^{3}}\approx8.92g/cm^{3}$.

Step6: Solve problem 5

Given $
ho = 10.5g/cm^{3}$ and $m = 2500.0g$, using $V=\frac{m}{
ho}$, we get $V=\frac{2500.0g}{10.5g/cm^{3}}\approx238.1cm^{3}$.

Step7: Solve problem 6

Given $
ho=0.8765g/mL$ and $V = 250.0mL$, using $m=
ho V$, we have $m=0.8765g/mL\times250.0mL=219.125g$.

Step8: Solve problem 7

First find volume $V$ of the lead - block. $V = 4.5cm\times5.2cm\times6.0cm=140.4cm^{3}$. Given $m = 1587g$, then $
ho=\frac{1587g}{140.4cm^{3}}\approx11.3g/cm^{3}$.

Step9: Solve problem 8

The volume of the iron shot is $V=49.10mL - 45.50mL=3.60mL$. Given $m = 28.5g$, then $
ho=\frac{28.5g}{3.60mL}\approx7.92g/mL$.

Answer:

1. $2.7g/mL$
2. $13.6g/mL$
3. $157.8g$
4. $\approx8.92g/cm^{3}$
5. $\approx238.1cm^{3}$
6. $219.125g$
7. $\approx11.3g/cm^{3}$
8. $\approx7.92g/mL$