Question

The density of seawater at $4^{\circ}\text{C}$ is}

• A. $1.008\text{ g/cm}^3$
• B. $1.0278\text{ g/cm}^3$
• C. $1.5000\text{ g/cm}^3$
• D. $2.0252\text{ g/cm}^3$

Hint:

Pure water density at $4^{\circ}\text{C}$ is exactly $1.0\text{ g/cm}^3$. Since seawater contains dissolved salts, its density must be slightly higher, making $1.0278\text{ g/cm}^3$ the most logical correct value.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Seawater density is determined by a combination of temperature, salinity, and pressure.
While pure water has a maximum density of $1.000\text{ g/cm}^3$ at $4^{\circ}\text{C}$, the presence of dissolved salts increases seawater density.
Key Formula or Approach:
The density of seawater (\(\rho\)) as a function of salinity (\(S\)), temperature (\(T\)), and pressure (\(P\)) is defined by the international equation of state of seawater.
For standard ocean salinity of approximately $35\text{ ppt}$ at $4^{\circ}\text{C}$ and atmospheric pressure:
\[ \rho \approx 1.0278\text{ g/cm}^3 \]

Step 2: Detailed Explanation:
Pure water reaches its peak density of $1.0000\text{ g/cm}^3$ at $4^{\circ}\text{C}$.
Seawater contains about $3.5\%$ dissolved salts (mostly sodium and chloride ions).
These dissolved salts add mass to a given volume of water without significantly increasing the volume, thereby raising its overall density.
At a standard temperature of $4^{\circ}\text{C}$, the density of seawater with $35\text{ ppt}$ salinity is approximately $1.0278\text{ g/cm}^3$ (or $1027.8\text{ kg/m}^3$).

Step 3: Final Answer:
Thus, the density of standard seawater at $4^{\circ}\text{C}$ is $1.0278\text{ g/cm}^3$.