Question

given the specific gravity of methyl alcohol 0.791, what is the mass of 2.50 l of methyl alcohol? a lab student measured out 85.00 g of benzene. given the specific gravity of benzene is 0.874, what is the volume of this solution in milliliters? a titanium (ti) cube has a mass of 605 mg. given the specific gravity of titanium is 4.51, what is the volume of the cube in cm³? a large piece of jewelry has a mass of 142.6 g. a graduated cylinder initially contains water and after adding the jewelry, the water level rises to 66.48 ml. assuming the jewelry is made of one substance and has a specific gravity of 10.49, what was the original volume of water in the graduated cylinder?

Explanation:

Problem 1: Mass of methyl alcohol

Step1: Define specific gravity relation

Specific gravity (SG) = $\frac{\text{Density of substance}}{\text{Density of water}}$, so density $
ho = \text{SG} \times
ho_{\text{water}}$. $
ho_{\text{water}} = 1000\ \frac{\text{g}}{\text{L}}$
$
ho = 0.791 \times 1000\ \frac{\text{g}}{\text{L}} = 791\ \frac{\text{g}}{\text{L}}$

Step2: Calculate mass from density

Mass $m =
ho \times V$, where $V=2.50\ \text{L}$
$m = 791\ \frac{\text{g}}{\text{L}} \times 2.50\ \text{L}$

Problem 2: Volume of benzene

Step1: Find density of benzene

$
ho = \text{SG} \times
ho_{\text{water}}$, $
ho_{\text{water}} = 1\ \frac{\text{g}}{\text{mL}}$
$
ho = 0.874 \times 1\ \frac{\text{g}}{\text{mL}} = 0.874\ \frac{\text{g}}{\text{mL}}$

Step2: Solve for volume

Volume $V = \frac{m}{
ho}$, where $m=85.00\ \text{g}$
$V = \frac{85.00\ \text{g}}{0.874\ \frac{\text{g}}{\text{mL}}}$

Problem 3: Volume of titanium cube

Step1: Convert mass to grams

$605\ \text{mg} = 0.605\ \text{g}$

Step2: Find density of titanium

$
ho = 4.51 \times 1\ \frac{\text{g}}{\text{cm}^3} = 4.51\ \frac{\text{g}}{\text{cm}^3}$

Step3: Calculate volume

$V = \frac{m}{
ho} = \frac{0.605\ \text{g}}{4.51\ \frac{\text{g}}{\text{cm}^3}}$

Problem 4: Original water volume

Step1: Find density of jewelry

$
ho = 10.49 \times 1\ \frac{\text{g}}{\text{mL}} = 10.49\ \frac{\text{g}}{\text{mL}}$

Step2: Calculate jewelry volume

$V_{\text{jewelry}} = \frac{m}{
ho} = \frac{142.6\ \text{g}}{10.49\ \frac{\text{g}}{\text{mL}}}$

Step3: Find original water volume

$V_{\text{original}} = V_{\text{final}} - V_{\text{jewelry}}$, where $V_{\text{final}}=66.48\ \text{mL}$

Answer:

1. Mass of methyl alcohol: $1980\ \text{g}$ (or $1.98\ \text{kg}$)
2. Volume of benzene: $97.25\ \text{mL}$
3. Volume of titanium cube: $0.134\ \text{cm}^3$
4. Original water volume: $52.70\ \text{mL}$