Question

part 1 of 2
a commercial rubbing alcohol contains 34. g of isopropanol and 0.194 g of sucrose octaacetate in each 45. ml portion. calculate the $\frac{weight}{volume}$ percent concentration for each component. be sure each of your answer entries has the correct number of significant figures.
$square$ % ($\frac{weight}{volume}$) isopropanol

Explanation:

Step1: Recall weight/volume % formula

The formula for weight/volume percent is $\% \frac{weight}{volume} = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100$

Step2: Plug in isopropanol values

Substitute mass = 34 g, volume = 45 mL
$\% \frac{weight}{volume} = \frac{34}{45} \times 100$

Step3: Calculate and round sig figs

$\frac{34}{45} \times 100 \approx 75.56$, rounded to 2 significant figures (from 34 and 45)

Answer:

76 % ($\frac{weight}{volume}$) isopropanol