Question

1. a flask that weighs 343.8 g is filled with 225 ml of carbon tetrachloride. the weight of the flask and carbon tetrachloride is found to be 701.55 g. from this information, calculate the density of carbon tetrachloride.
2. calculate the density of sulfuric acid if 35.4 ml of the acid weighs 65.14 g.
3. find the mass of 250.0 ml of benzene. the density of benzene is 0.8765 g/ml.
4. a block of lead has dimensions of 4.50 cm by 5.20 cm by 6.00 cm. the block weighs 1587 g. from this information, calculate the density of lead.
5. 28.5 g of iron is added to a graduated cylinder containing 45.50 ml of water. the water level rises to the 49.10 ml mark. from this information, calculate the density of iron.
6. what volume of silver metal will weigh exactly 2500.0 g. the density of silver is 10.5 g/cm³

Explanation:

Step1: Recall density formula

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

5)

• First, find the mass of carbon - tetrachloride:
• The mass of the flask is $m_{flask}=345.8\ g$ and the mass of the flask + carbon - tetrachloride is $m_{total}=701.55\ g$.
• So the mass of carbon - tetrachloride $m = m_{total}-m_{flask}=701.55 - 345.8=355.75\ g$.
• The volume of carbon - tetrachloride $V = 225\ mL$.
• Using the density formula $

ho=\frac{m}{V}$, we have $
ho=\frac{355.75\ g}{225\ mL}\approx1.581\ g/mL$.

6)

• Given $m = 65.14\ g$ and $V = 35.4\ mL$.
• Using the density formula $

ho=\frac{m}{V}=\frac{65.14\ g}{35.4\ mL}\approx1.84\ g/mL$.

7)

• Given $

ho = 0.8765\ g/mL$ and $V = 250.0\ mL$.

• Using the formula $m=

ho V$, we get $m=(0.8765\ g/mL)\times250.0\ mL = 219.125\ g$.

8)

• First, find the volume of the lead block:
• The volume of a rectangular - shaped block with length $l = 4.50\ cm$, width $w = 5.20\ cm$ and height $h = 6.00\ cm$ is $V=l\times w\times h=(4.50\ cm)\times(5.20\ cm)\times(6.00\ cm)=140.4\ cm^{3}$.
• The mass of the block $m = 1587\ g$.
• Using the density formula $

ho=\frac{m}{V}=\frac{1587\ g}{140.4\ cm^{3}}\approx11.3\ g/cm^{3}$.

9)

• First, find the volume of iron:
• The initial volume of water $V_1 = 45.50\ mL$ and the final volume of water + iron $V_2 = 49.10\ mL$.
• So the volume of iron $V=V_2 - V_1=49.10\ mL - 45.50\ mL = 3.60\ mL$.
• The mass of iron $m = 28.5\ g$.
• Using the density formula $

ho=\frac{m}{V}=\frac{28.5\ g}{3.60\ mL}\approx7.92\ g/mL$.

10)

• Given $m = 2500.0\ g$ and $

ho = 10.5\ g/cm^{3}$.

• Using the formula $V=\frac{m}{

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

Answer:

1. $

ho\approx1.581\ g/mL$

1. $

ho\approx1.84\ g/mL$

1. $m = 219.125\ g$
2. $

ho\approx11.3\ g/cm^{3}$

1. $

ho\approx7.92\ g/mL$

1. $V\approx238.1\ cm^{3}$