Based on the slope of the line between 45°C (80 g KBr/100 g water) and 90°C (100 g KBr/100 g water), the interpolation suggests approximately 77.8 g of potassium bromide can dissolve in 100 g of water at 40°C.
To determine the amount of potassium bromide (KBr) that can be dissolved in 100 grams of water at 40°C, we can interpolate based on the information provided by the graph.
1. Identify the coordinates for the relevant data points on the line:
- Point A: (45°C, 80 g KBr/100 g water)
- Point B: (90°C, 100 g KBr/100 g water)
2. Calculate the slope (m) of the line using the formula:
![\[ m = \frac{\text{Change in } y}{\text{Change in } x} = \frac{\text{(100 - 80)}}{\text{(90 - 45)}} \]](https://img.qammunity.org/2024/formulas/biology/college/fnor9qjmonien5zwe7u03yxll4b9ft9x18.png)
3. Use the calculated slope to find the amount of KBr (y) dissolved in 100 grams of water at 40°C:
![\[ y - 80 = m * (40 - 45) \] \[ y = 80 + m * (-5) \]](https://img.qammunity.org/2024/formulas/biology/college/slxesj4uq1ol8v792j1bmfy31rpq27jyvx.png)
Substitute the calculated slope (m) into the equation to solve for y.
4. Calculate the value of y:
![\[ y = 80 - \left((20)/(45)\right) * 5 \]](https://img.qammunity.org/2024/formulas/biology/college/35o1ibb6re55dpmoav5oizwcp3gg4strql.png)
5. Solve for y to find the amount of KBr at 40°C:
![\[ y \approx 77.78 \, \text{g} \]](https://img.qammunity.org/2024/formulas/biology/college/9gqpd1vynn21d19rowqur9hgdn44r81rh2.png)
Therefore, according to graph, approximately 77.78 grams of potassium bromide can be dissolved in 100 grams of water at 40°C.
The complete question is:
According to graph, how much potassium bromide can be dissolved in 100 g of water at 40°C?