Question

What is the percent-by-mass, %(m/m), concentration of sucrose ( table sugar) in a solution made by dissolving 22.8 g of sucrose in 41.7 g of water

Answer 1

Answer:

$\%m/m=35.3\%$

Explanation:

Hello,

In this case, we understand the by-mass percentage as a relationship between the mass of the solute (in this case sucrose) and the mass of the solution (composed by the sucrose and the water):

$\%m/m=\frac{m{solute}}{m_{solution}}*100\%$

Thus, since the solution is composed both both the sucrose and the water, the required percentage results in:

$\%m/m=\frac{22.8g}{41.7g+22.8g}*100\%\\\\\%m/m=35.3\%$

Best regards.

Answer 2

Answer:

32.6 %

Explanation:

Given data

• Mass of sucrose (solute): 22.8 grams

• Mass of water (solvent): 47.1 grams

Step 1: Calculate the mass of the solution

The mass of the solution is equal to the sum of the mass of the solute and the mass of the solvent.

m(solution) = m(solute) + m(solvent)

m(solution) = 22.8 g + 47.1 g

m(solution) = 69.9 g

Step 2: Calculate the percent-by-mass of sucrose in the solution

We will use the following expression.

$\% m/m = \frac{mass\ of\ sucrose}{mass\ of\ solution} \times 100\% = \frac{22.8g}{69.9g} \times 100\% = 32.6 \%$