QUESTION IMAGE
Question
1.6 specific gravity
explain the difference between density and specific gravity.
if given the density of a substance is 3.00 g/cm³, what would the specific gravity for that substance be?
if given the density of a substance is 285 lbs/ft³ and the density of water is 62.416 lbs/ft³, what is the specific gravity of the substance?
practice:
a student measured out 25.0 ml of a saline (salt water) solution. what is the mass of the solution, given the specific gravity of the solution is 1.09.
an irregularly shaped stone was lowered into a graduated cylinder holding a volume of water equal to 6.03 ml. the height of the water rose to 9.17 ml. if the mass of the stone is 15.292 g, calculate the specific gravity of the stone.
Step1: Define density vs specific gravity
Density is mass per unit volume ($
ho = \frac{m}{V}$) with units (e.g., $\text{g/cm}^3$, $\text{lbs/ft}^3$). Specific gravity (SG) is the ratio of a substance's density to pure water's density at the same conditions, so it is unitless: $\text{SG} = \frac{
ho_{\text{substance}}}{
ho_{\text{water}}}$.
---
Step2: Calculate SG for 3.00 g/cm³
Water density at standard conditions is $1.00\ \text{g/cm}^3$.
$\text{SG} = \frac{3.00\ \text{g/cm}^3}{1.00\ \text{g/cm}^3}$
Step3: Solve for SG
$\text{SG} = 3.00$
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Step4: Calculate SG for 285 lbs/ft³
Use given water density $
ho_{\text{water}} = 62.416\ \text{lbs/ft}^3$.
$\text{SG} = \frac{285\ \text{lbs/ft}^3}{62.416\ \text{lbs/ft}^3}$
Step5: Solve for SG
$\text{SG} \approx 4.57$
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Step6: Find saline solution mass
First, use SG to get solution density: $
ho_{\text{solution}} = \text{SG} \times
ho_{\text{water}}$. Since $1\ \text{mL water} = 1\ \text{g}$, $
ho_{\text{water}} = 1\ \text{g/mL}$.
$
ho_{\text{solution}} = 1.09 \times 1\ \text{g/mL} = 1.09\ \text{g/mL}$
Step7: Calculate mass from density
$m =
ho \times V = 1.09\ \text{g/mL} \times 25.0\ \text{mL}$
Step8: Solve for mass
$m = 27.25\ \text{g}$
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Step9: Find stone volume
Volume of stone = final water volume - initial water volume.
$V_{\text{stone}} = 9.17\ \text{mL} - 6.03\ \text{mL} = 3.14\ \text{mL}$
Since $1\ \text{mL} = 1\ \text{cm}^3$, $V_{\text{stone}} = 3.14\ \text{cm}^3$
Step10: Calculate stone density
$
ho_{\text{stone}} = \frac{m}{V} = \frac{15.292\ \text{g}}{3.14\ \text{cm}^3}$
$
ho_{\text{stone}} \approx 4.87\ \text{g/cm}^3$
Step11: Calculate stone SG
$\text{SG} = \frac{4.87\ \text{g/cm}^3}{1.00\ \text{g/cm}^3}$
Step12: Solve for SG
$\text{SG} \approx 4.87$
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- Difference between density and specific gravity: Density is a measure of mass per unit volume of a substance (has units like $\text{g/cm}^3$). Specific gravity is a unitless ratio of a substance's density to the density of pure water at the same temperature and pressure.
- Specific gravity of the $3.00\ \text{g/cm}^3$ substance: $3.00$
- Specific gravity of the $285\ \text{lbs/ft}^3$ substance: $\approx 4.57$
- Mass of the saline solution: $27.3\ \text{g}$ (rounded to 3 significant figures)
- Specific gravity of the stone: $\approx 4.87$