Question

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a sailor on a trans - pacific solo voyage notices one day that if he puts 758. 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 mass 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.
youll need to know that the density of fresh water at the temperature of the sea around the sailor is 0.999 g/cm³. youll also want to remember archimedes principle, that objects float when they displace a mass of water equal to their own mass.
35 g

Explanation:

Step1: Calculate mass of fresh water

Density formula: $
ho = \frac{m}{V}$, so $m =
ho V$.
Fresh water volume $V = 758\ \text{mL} = 758\ \text{cm}^3$ (since $1\ \text{mL} = 1\ \text{cm}^3$), density $
ho = 0.999\ \text{g/cm}^3$.
$m_{\text{water}} = 0.999\ \text{g/cm}^3 \times 758\ \text{cm}^3 = 757.242\ \text{g}$.

Step2: Total mass of cup and water

$m_{\text{total}} = m_{\text{cup}} + m_{\text{water}} = 25.0\ \text{g} + 757.242\ \text{g} = 782.242\ \text{g}$.

Step3: Mass of displaced seawater (Archimedes' Principle)

Displaced seawater mass equals total mass: $m_{\text{seawater}} = 782.242\ \text{g}$.
Volume of displaced seawater is equal to volume of fresh water (since water levels are equal): $V_{\text{seawater}} = 758\ \text{cm}^3 = 0.758\ \text{L}$.

Step4: Density of seawater

$
ho_{\text{seawater}} = \frac{m_{\text{seawater}}}{V_{\text{seawater}}} = \frac{782.242\ \text{g}}{758\ \text{cm}^3} \approx 1.032\ \text{g/cm}^3$.

Step5: Mass of salt per liter

Density of fresh water is $0.999\ \text{g/cm}^3 = 999\ \text{g/L}$.
Density of seawater is $1.032\ \text{g/cm}^3 = 1032\ \text{g/L}$.
Mass of salt per liter: $1032\ \text{g/L} - 999\ \text{g/L} = 33\ \text{g/L}$ (rounded to 2 sig figs: $33\ \text{g/L}$ or check calculation again—wait, maybe miscalculation. Wait, step3: displaced seawater volume is same as cup's submerged volume? Wait, no—wait, the cup has fresh water, so the displaced seawater volume is the volume of the cup plus fresh water? No, the problem says "fresh water inside the cup at exactly the same level as the seawater outside"—so the volume of displaced seawater is equal to the volume of the fresh water (since the cup's submerged volume plus fresh water volume? No, no—when the water levels are equal, the volume of displaced seawater is equal to the volume of the fresh water (because the fresh water is at the same level as seawater, so the submerged volume of the cup plus fresh water volume equals displaced seawater volume? Wait, no—actually, the cup floats, so the displaced seawater volume is equal to the volume of the cup that is submerged plus the volume of fresh water? No, the fresh water is inside the cup, so the total displaced volume is the volume of the cup below the water level plus the volume of fresh water. But since the fresh water level is equal to seawater level, the volume of fresh water is equal to the volume of the cup's submerged part? No, maybe simpler: the mass of displaced seawater is equal to the total mass (cup + fresh water). The volume of displaced seawater is equal to the volume of fresh water (because the fresh water is at the same level, so the volume of seawater displaced is equal to the volume of fresh water, since the cup's own volume displacement plus fresh water volume? Wait, no—let's re-express.

Wait, the fresh water is in the cup, so the total mass is cup + fresh water. The displaced seawater mass is equal to that total mass (Archimedes). The volume of displaced seawater is the volume of the cup that is submerged plus the volume of fresh water. But since the fresh water level is equal to seawater level, the volume of fresh water is equal to the volume of the cup's submerged part? No, actually, the key is that the volume of displaced seawater is equal to the volume of the fresh water (because the fresh water is at the same level, so the amount of seawater displaced is equal to the volume of fresh water, because the cup's own volume displacement is canceled by the fresh water? Wait, maybe I made a mistake earlier. Let's redo step3 and 4.

Wait, the fresh water is 758 mL, so the volume of the cup that is submerged plus 758 mL equals the displaced seawater volume. But since the fresh water level is equal to seawater level, the submerged volume of the cup is equal to the volume of the cup above the fresh water? No, this is confusing. Wait, the problem says "fresh water inside the cup at exactly the same level as the seawater outside the cup"—so the total volume of displaced seawater is equal to the volume of the fresh water (because the cup's wall is thin, so the volume of the cup itself is negligible? Or maybe the volume of displaced seawater is equal to the volume of fresh water, because the water levels are equal, so the submerged volume of the cup plus fresh water volume equals displaced seawater volume, but since the cup is floating, the submerged volume of the cup times seawater density equals the total mass. Wait, no—Archimedes' Principle: the buoyant force equals the weight of displaced fluid. The buoyant force must equal the weight of the cup plus fresh water. So:

$F_b = m_{\text{total}} g =
ho_{\text{seawater}} V_{\text{displaced}} g$

Cancel $g$: $m_{\text{total}} =
ho_{\text{seawater}} V_{\text{displaced}}$.

But $V_{\text{displaced}}$ is the volume of seawater displaced, which is equal to the volume of the cup that is submerged plus the volume of fresh water. But since the fresh water is at the same level as seawater, the volume of fresh water is equal to the volume of the cup's submerged part? No, that can't be. Wait, the fresh water is inside the cup, so the cup has a certain volume, and the fresh water fills it up to the seawater level. So the total volume displaced is the volume of the cup (submerged) plus the volume of fresh water. But the fresh water volume is 758 mL, so the submerged volume of the cup is zero? No, the cup is floating, so part of it is submerged. Wait, maybe the problem is simplified: the volume of displaced seawater is equal to the volume of fresh water (758 mL), because the fresh water is at the same level as seawater, so the cup's submerged volume is equal to the volume of the cup above the water, but that's not right. Wait, let's go back to the problem statement: "fresh water inside the cup at exactly the same level as the seawater outside the cup"—so the top of the fresh water is at the same level as the seawater surface. So the volume of the cup that is submerged is equal to the volume of the cup below the water level, and the fresh water is inside the cup, so the total displaced volume is the submerged volume of the cup plus the volume of fresh water. But since the cup is floating, the buoyant force equals the weight of the cup plus fresh water. The buoyant force is also equal to the weight of displaced seawater, which is $
ho_{\text{seawater}} \times V_{\text{displaced}} \times g$. The weight of the cup plus water is $(m_{\text{cup}} + m_{\text{water}})g$. So:

$
ho_{\text{seawater}} \times V_{\text{displaced}} = m_{\text{cup}} + m_{\text{water}}$.

Now, what is $V_{\text{displaced}}$? The fresh water is at the same level as seawater, so the volume of the cup that is submerged is equal to the volume of the cup below the water level, and the fresh water is inside the cup, so the total volume displaced is the submerged volume of the cup plus the volume of fresh water. But the fresh water volume is 758 mL, so the submerged volume of the cup is $V_{\text{cup, submerged}}$, and $V_{\text{displaced}} = V_{\text{cup, submerged}} + 758\ \text{mL}$. But we don't know $V_{\text{cup, submerged}}$. Wait, maybe the cup's volume is negligible, or the problem assumes that the volume of displaced seawater is equal to the volume of fresh water (758 mL). Let's check the initial calculation again.

Wait, the user's initial answer was 35 g, but let's recalculate:

Mass of fresh water: $0.999\ \text{g/mL} \times 758\ \text{mL} = 757.242\ \text{g}$.
Total mass: $25 + 757.242 = 782.242\ \text{g}$.
Volume of displaced seawater: 758 mL (since water levels are equal, so the volume of seawater displaced is equal to the volume of fresh water, because the cup's own volume displacement is canceled by the fresh water? No, that doesn't make sense. Wait, maybe the key is that the volume of the fresh water is equal to the volume of the seawater displaced, because the water levels are the same. So $V_{\text{seawater}} = 758\ \text{mL} = 0.758\ \text{L}$.

Then, density of seawater: $782.242\ \text{g} / 758\ \text{mL} = 1.032\ \text{g/mL}$.

Density of fresh water: $0.999\ \text{g/mL}$.

Mass of salt per liter: $(1.032 - 0.999)\ \text{g/mL} \times 1000\ \text{mL/L} = 33\ \text{g/L}$ (rounded to 2 sig figs: 33 g/L, or 33 g per liter). Wait, but maybe my mistake was in the volume of displaced seawater. Let's check again.

Wait, the cup has mass 25 g, fresh water 758 mL (757.242 g), total 782.242 g. The displaced seawater must have mass 782.242 g. The volume of displaced seawater is the volume of the cup plus fresh water? No, the fresh water is inside the cup, so the cup's volume is, say, $V_{\text{cup}}$, and the fresh water is 758 mL, so the total volume displaced is $V_{\text{cup, submerged}} + 758\ \text{mL}$. But we don't know $V_{\text{cup, submerged}}$. Wait, maybe the cup's volume is negligible, so $V_{\text{displaced}} \approx 758\ \text{mL}$. Then density of seawater is 782.242 g / 758 mL ≈ 1.032 g/mL. Then mass of salt per liter is (1.032 - 0.999) g/mL * 1000 mL/L = 33 g/L. Rounded to 2 sig figs, 33 g/L, or 33 g per liter. Wait, but maybe the correct answer is 33 g/L, or 33 g. Wait, the initial answer was 35 g, which is wrong. Let's recalculate:

Wait, 758 mL fresh water: mass = 0.999 * 758 = 757.242 g.
Cup: 25 g. Total: 782.242 g.
Displaced seawater: 782.242 g, volume 758 mL (since water levels are equal, so the volume of seawater displaced is equal to the volume of fresh water, because the cup's submerged volume is equal to the volume of the cup above the water, but that's not correct. Wait, no—when the water levels are equal, the volume of the fresh water is equal to the volume of the seawater displaced by the cup (excluding the fresh water). Wait, this is confusing. Let's use Archimedes' Principle correctly: the buoyant force equals the weight of the displaced fluid. The object (cup + fresh water) floats, so buoyant force equals its weight. The displaced fluid is seawater, so:

$m_{\text{cup}} + m_{\text{water}} = m_{\text{seawater displaced}}$.

The volume of seawater displaced is equal to the volume of the cup that is submerged plus the volume of the fresh water. But since the fresh water is at the same level as seawater, the volume of the cup that is submerged is equal to the volume of the cup below the water level, and the fresh water fills the cup up to the water level, so the volume of the cup is equal to the volume of fresh water (758 mL) plus the volume of the cup above the water level. But the cup is floating, so the submerged volume of the cup times seawater density equals the weight of the cup. Wait, no—let's split the forces: the buoyant force on the cup is $
ho_{\text{seawater}} \times V_{\text{cup, submerged}} \times g$, and the buoyant force on the fresh water is $
ho_{\text{seawater}} \times V_{\text{water}} \times g$ (since the fresh water is at the same level, so it's like a column of seawater, but no—fresh water is less dense, so it would float, but it's inside the cup. Wait, the fresh water is in the cup, so the cup is holding the fresh water, and the entire system (cup + fresh water) floats. So the total buoyant force is the weight of the displaced seawater, which is equal to the weight of the cup plus fresh water. The volume of displaced seawater is equal to the volume of the cup that is submerged plus the volume of the fresh water. But the fresh water is at the same level as seawater, so the volume of the cup that is submerged is equal to the volume of the cup below the water level, and the fresh water fills the cup up to the water level, so the volume of the cup is $V_{\text{cup}} = V_{\text{cup, submerged}} + V_{\text{cup, above}}$. The fresh water volume is $V_{\text{water}} = V_{\text{cup, above}}$, because it fills the cup up to the water level. Therefore, $V_{\text{cup, submerged}} = V_{\text{cup}} - V_{\text{water}}$, and the total displaced volume is $V_{\text{cup, submerged}} + V_{\text{water}} = V_{\text{cup}}$. But we don't know $V_{\text{cup}}$. This is a problem. Wait, maybe the problem assumes that the volume of the cup is negligible, so $V_{\text{displaced}} \approx V_{\text{water}} = 758\ \text{mL}$. Then:

$m_{\text{seawater displaced}} = 782.242\ \text{g}$, $V_{\text{displaced}} = 758\ \text{mL}$, so density of seawater is $782.242 / 758 ≈ 1.032\ \text{g/mL}$.

Density of fresh water is $0.999\ \text{g/mL}$, so the mass of salt per mL is $1.032 - 0.999 = 0.033\ \text{g/mL}$.

Mass per liter: $0.033\ \text{g/mL} \times 1000\ \text{mL/L} = 33\ \text{g/L}$.

Rounded to 2 significant figures, that's 33 g/L, or 33 g per liter. So the correct answer should be 33 g/L (or 33 g, but with unit g/L). Wait, the problem says "mass of salt dissolved in each liter of seawater", so unit is g/L. So 33 g/L, or 33 g (if unit is implied as per liter). Wait, maybe my initial calculation had a mistake in the volume. Let's check again:

$m_{\text{water}} = 0.999 * 758 = 757.242\ \text{g}$.
$m_{\text{total}} = 25 + 757.242 = 782.242\ \text{g}$.
$V_{\text{seawater}} = 758\ \text{mL} = 0.758\ \text{L}$.
Density of seawater: $782.242\ \text{g} / 0.758\ \text{L} = 1032\ \text{g/L}$.
Density of fresh water: $0.999\ \text{g/mL} = 999\ \text{g/L}$.
Mass of salt per liter: $1032 - 999 = 33\ \text{g/L}$.

Yes, that's correct. So the mass of salt per liter is 33 g (rounded to 2 sig figs).

Answer:

33 g/L (or 33 g, with the understanding that it's per liter)