Question

a commercial rubbing alcohol contains 32. g of isopropanol and 0.250 g of sucrose octaacetate in each 52. ml portion. calculate the weight/volume percent concentration for each component. be sure each of your answer entries has the correct number of significant figures.

Explanation:

Step1: Calculate the weight - volume percent for isopropanol

The formula for weight - volume percent (\(\%\text{ w/v}\)) is \(\%\text{ w/v}=\frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}}\times100\).
Given the mass of isopropanol \(m = 32\space g\) and the volume of the solution \(V=52\space mL\).
Substitute the values into the formula: \(\%\text{ w/v}=\frac{32\space g}{52\space mL}\times 100\).

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Rounding to two significant figures (since \(32\) has two significant figures and \(52\) has two significant figures), we get \(62\%\text{ w/v}\).

Step2: Calculate the weight - volume percent for sucrose octaacetate

Given the mass of sucrose octaacetate \(m = 0.250\space g\) and the volume of the solution \(V = 52\space mL\).
Using the formula \(\%\text{ w/v}=\frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}}\times100\).
Substitute the values: \(\%\text{ w/v}=\frac{0.250\space g}{52\space mL}\times100\).

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Rounding to two significant figures (since \(0.250\) has three significant figures and \(52\) has two significant figures, we go by the least number of significant figures from the given values in the calculation, which is two), we get \(0.48\%\text{ w/v}\)

Answer:

The weight - volume percent of isopropanol is \(62\%\text{ w/v}\) and the weight - volume percent of sucrose octaacetate is \(0.48\%\text{ w/v}\)