The vapor pressure of water at 25.0°C is 23.8 torr. Determine the mass of 180 g/mol) needed to add to 500.0 g of water to glucose (molar mass = change the vapor pressure to 23.5 torr. (Ignore significant figures for this problem.) O 6.4 g O 64 g O 178 g O 6.4 kg 182 g

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### Determining the Mass of Glucose Needed to Change Water's Vapor Pressure

The vapor pressure of water at 25.0°C is 23.8 torr. To solve this problem, we need to determine the mass of glucose that should be added to 500.0 g of water to adjust its vapor pressure to 23.5 torr. The molar mass of glucose is given as 180 g/mol. For this calculation, we will ignore significant figures.

#### Multiple-Choice Options
- ○ 6.4 g
- ○ 64 g
- ○ 178 g
- ○ 6.4 kg
- ● 182 g

*(The correct option is indicated with a blue dot.)*

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This example can be used to understand the colligative property of solutions, specifically how the addition of a non-volatile solute (glucose) affects the vapor pressure of a solvent (water).
Transcribed Image Text:--- ### Determining the Mass of Glucose Needed to Change Water's Vapor Pressure The vapor pressure of water at 25.0°C is 23.8 torr. To solve this problem, we need to determine the mass of glucose that should be added to 500.0 g of water to adjust its vapor pressure to 23.5 torr. The molar mass of glucose is given as 180 g/mol. For this calculation, we will ignore significant figures. #### Multiple-Choice Options - ○ 6.4 g - ○ 64 g - ○ 178 g - ○ 6.4 kg - ● 182 g *(The correct option is indicated with a blue dot.)* --- This example can be used to understand the colligative property of solutions, specifically how the addition of a non-volatile solute (glucose) affects the vapor pressure of a solvent (water).
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