Question

what is the density of a sample with specific gravity = 1.0999 at 28.0 °c if the density of water at 28.0 °c is 0.9961 g/cm³?

d_sample

the relationship is sp gr = ---------- and we need to solve for 1.

d_water

2 = (0.9961 g/cm³)(3) = 4

a. d_sample b. d_water c. sp gr d. 1.096 e. 1.104 f. 1.096 g/cm³

g. 1.104 g/cm³

Explanation:

Step1: Recall density - specific gravity formula

The formula for specific gravity (sp gr) is $sp\ gr=\frac{D_{sample}}{D_{water}}$, and we need to solve for $D_{sample}$.

Step2: Rearrange the formula

Since $sp\ gr=\frac{D_{sample}}{D_{water}}$, then $D_{sample}=(D_{water})\times(sp\ gr)$.

Step3: Substitute given values

We know $D_{water} = 0.9961\ g/cm^{3}$ and $sp\ gr=1.0999$. So $D_{sample}=(0.9961\ g/cm^{3})\times(1.0999)\approx1.096\ g/cm^{3}$.

Answer:

1. A. $D_{sample}$
2. $D_{sample}$
3. $1.0999$
4. $1.096\ g/cm^{3}$