QUESTION 23 Using the calibration curve below (equation of the line: y=0.0039x + 0.9981) to determine the percentage of sugar in a sugar water solution with a density of 1.034 g/mL. Density versus Percent Sugar 1.07 1.06 y = 0.0039x + 0.9981 R= 0.9998 1.05 1.04 1.03 1.02 1.01 0.99 10 12 14 16 Percent Sugar (% m/m ) O 4.2 % O 1.0% O 8.5% 9.9% O 9.2% Density (g/mL)

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### Question 23

**Using the calibration curve below (equation of the line: y=0.0039x + 0.9981) to determine the percentage of sugar in a sugar water solution with a density of 1.034 g/mL.**

### Graph Explanation

The graph shown is titled "Density versus Percent Sugar." It depicts the relationship between the density of a sugar water solution (y-axis) and the percentage of sugar in the solution (x-axis).

- **x-axis (independent variable):** Percentage of sugar (% m/m) ranging from 0% to 15%.
- **y-axis (dependent variable):** Density (g/mL) ranging from 0.99 g/mL to 1.07 g/mL.

Four data points are plotted on the graph, and a linear trendline is fitted to these points. The equation of the trendline is provided as:

\[ y = 0.0039x + 0.9981 \]

The R² value of 0.9998 indicates a very high correlation between density and percentage of sugar, implying that the linear model is an excellent fit for the data.

### Calculation

To determine the percentage of sugar in a solution with a given density (1.034 g/mL), use the equation of the line provided:

\[ y = 0.0039x + 0.9981 \]

Given that y (density) = 1.034 g/mL, solve for x (percentage of sugar):

\[ 1.034 = 0.0039x + 0.9981 \]

Subtract 0.9981 from both sides:

\[ 1.034 - 0.9981 = 0.0039x \]

\[ 0.0359 = 0.0039x \]

Solve for x by dividing both sides by 0.0039:

\[ x = \frac{0.0359}{0.0039} \]

\[ x = 9.2 \% m/m \]

Therefore, a sugar water solution with a density of 1.034 g/mL contains approximately 9.2% sugar by mass.

**Answer Choices:**
- 4.2%
- 1.0%
- 8.5%
- 9.9%
- 9.2%

Based on the calculation, the correct answer is **9.2%**
Transcribed Image Text:### Question 23 **Using the calibration curve below (equation of the line: y=0.0039x + 0.9981) to determine the percentage of sugar in a sugar water solution with a density of 1.034 g/mL.** ### Graph Explanation The graph shown is titled "Density versus Percent Sugar." It depicts the relationship between the density of a sugar water solution (y-axis) and the percentage of sugar in the solution (x-axis). - **x-axis (independent variable):** Percentage of sugar (% m/m) ranging from 0% to 15%. - **y-axis (dependent variable):** Density (g/mL) ranging from 0.99 g/mL to 1.07 g/mL. Four data points are plotted on the graph, and a linear trendline is fitted to these points. The equation of the trendline is provided as: \[ y = 0.0039x + 0.9981 \] The R² value of 0.9998 indicates a very high correlation between density and percentage of sugar, implying that the linear model is an excellent fit for the data. ### Calculation To determine the percentage of sugar in a solution with a given density (1.034 g/mL), use the equation of the line provided: \[ y = 0.0039x + 0.9981 \] Given that y (density) = 1.034 g/mL, solve for x (percentage of sugar): \[ 1.034 = 0.0039x + 0.9981 \] Subtract 0.9981 from both sides: \[ 1.034 - 0.9981 = 0.0039x \] \[ 0.0359 = 0.0039x \] Solve for x by dividing both sides by 0.0039: \[ x = \frac{0.0359}{0.0039} \] \[ x = 9.2 \% m/m \] Therefore, a sugar water solution with a density of 1.034 g/mL contains approximately 9.2% sugar by mass. **Answer Choices:** - 4.2% - 1.0% - 8.5% - 9.9% - 9.2% Based on the calculation, the correct answer is **9.2%**
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