A sailor on a trans-Pacific solo voyage notices one day that if he puts 714. mL of fresh water into a plastic cup weighing 25.0 g, the cup floats in the seawater around his boat with the fresh water inside the cup at exactly fresh water the same level as the seawater outside the cup (see sketch at right). salt water Calculate the amount of salt dissolved in each liter of seawater. Be sure your answer has a unit symbol, if needed, and round it to 2 significant digits. You'll need to know that the density of fresh water at the temperature of the sea around the sailor is 0.999 g/cm. You'll also want to remember Archimedes' Principle, that objects float when they displace a mass of water equal to their own mass.

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### Example Problem: Calculating the Mass of Dissolved Salt in Seawater

A sailor on a trans-Pacific solo voyage notices one day that if he puts 714 mL of fresh water into a plastic cup weighing 25.0 g, the cup floats in the seawater around his boat with the fresh water inside the cup at exactly the same level as the seawater outside the cup (*see sketch at right*).

Calculate the amount of salt dissolved in each liter of seawater. Be sure your answer has a unit symbol, if needed, and round it to 2 significant digits.

#### Key Information
1. The density of fresh water at the temperature of the sea around the sailor is \(0.999 \text{ g/cm}^3\).
2. Archimedes' Principle: Objects float when they displace a mass of water equal to their own mass.

#### Steps to Solve
1. **Determine the mass of the freshwater inside the cup**:
   - Volume of fresh water = 714 mL
   - Density of fresh water \(= 0.999 \text{ g/cm}^3 \)
   - Convert volume to cm\(^3\) (since 1 mL = 1 cm\(^3\)):
     \[
     714 \text{ mL} = 714 \text{ cm}^3
     \]
   - Mass of fresh water:
     \[
     \text{Mass} = \text{Volume} \times \text{Density}  = 714 \text{ cm}^3 \times 0.999 \text{ g/cm}^3 = 712.29 \text{ g}
     \]

2. **Total mass floating in seawater**:
   - Mass of cup \(= 25.0 \text{ g}\)
   - Total mass = Mass of cup + Mass of fresh water
     \[
     \text{Total mass} = 25.0 \text{ g} + 712.29 \text{ g} = 737.29 \text{ g}
     \]

3. **Apply Archimedes' Principle**:
   - The seawater displaced by the cup with fresh water inside is equal to the total mass floating, which is 737.29 g.

4. **Calculate the density of seawater**:
   - Let \(\rho_s\) be the density of
Transcribed Image Text:### Example Problem: Calculating the Mass of Dissolved Salt in Seawater A sailor on a trans-Pacific solo voyage notices one day that if he puts 714 mL of fresh water into a plastic cup weighing 25.0 g, the cup floats in the seawater around his boat with the fresh water inside the cup at exactly the same level as the seawater outside the cup (*see sketch at right*). Calculate the amount of salt dissolved in each liter of seawater. Be sure your answer has a unit symbol, if needed, and round it to 2 significant digits. #### Key Information 1. The density of fresh water at the temperature of the sea around the sailor is \(0.999 \text{ g/cm}^3\). 2. Archimedes' Principle: Objects float when they displace a mass of water equal to their own mass. #### Steps to Solve 1. **Determine the mass of the freshwater inside the cup**: - Volume of fresh water = 714 mL - Density of fresh water \(= 0.999 \text{ g/cm}^3 \) - Convert volume to cm\(^3\) (since 1 mL = 1 cm\(^3\)): \[ 714 \text{ mL} = 714 \text{ cm}^3 \] - Mass of fresh water: \[ \text{Mass} = \text{Volume} \times \text{Density} = 714 \text{ cm}^3 \times 0.999 \text{ g/cm}^3 = 712.29 \text{ g} \] 2. **Total mass floating in seawater**: - Mass of cup \(= 25.0 \text{ g}\) - Total mass = Mass of cup + Mass of fresh water \[ \text{Total mass} = 25.0 \text{ g} + 712.29 \text{ g} = 737.29 \text{ g} \] 3. **Apply Archimedes' Principle**: - The seawater displaced by the cup with fresh water inside is equal to the total mass floating, which is 737.29 g. 4. **Calculate the density of seawater**: - Let \(\rho_s\) be the density of
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