Coefficient of apparent expansions of mercury is `0.18 xx 10^(-3)//^(0)C`. If the density of mercury at `0^(0)C` is `13.6 g//c c` its density at `473K` will be

Options

A. `13.11g//c c`
B. `26.22g//c c`
C. `52.11g//c c`
D. None of these

Explanation

To find the density of mercury at 473 K, we can use the formula for the density change with temperature, which is given by:

\[ \rho = \rho_0 (1 - \gamma \Delta \theta) \]

Where:
- \(\rho\) is the density at the new temperature,
- \(\rho_0\) is the initial density at \(0^\circ C\),
- \(\gamma\) is the coefficient of apparent expansion,
- \(\Delta \theta\) is the change in temperature in degrees Celsius.

### Step 1: Identify the given values
- Coefficient of apparent expansion, \(\gamma = 0.18 \times 10^{-3} \, ^\circ C^{-1}\)
- Density at \(0^\circ C\), \(\rho_0 = 13.6 \, g/cm^3\)
- Initial temperature, \(T_0 = 0^\circ C\) (which is equivalent to \(273 K\))
- Final temperature, \(T_f = 473 K\)

### Step 2: Calculate the change in temperature (\(\Delta \theta\))
\[ \Delta \theta = T_f - T_0 = 473 K - 273 K = 200 K \]
Since we are using degrees Celsius, \(\Delta \theta = 200^\circ C\).

### Step 3: Substitute the values into the density formula
Substituting the known values into the formula:
\[ \rho = 13.6 \, g/cm^3 \times \left(1 - (0.18 \times 10^{-3}) \times 200\right) \]

### Step 4: Calculate the term inside the parentheses
Calculating the product:
\[ 0.18 \times 10^{-3} \times 200 = 0.036 \]
Now substituting this back:
\[ \rho = 13.6 \, g/cm^3 \times (1 - 0.036) \]
\[ \rho = 13.6 \, g/cm^3 \times 0.964 \]

### Step 5: Perform the final multiplication
Calculating the final density:
\[ \rho = 13.6 \times 0.964 \approx 13.11344 \, g/cm^3 \]

### Step 6: Round the answer
Rounding to two decimal places, we get:
\[ \rho \approx 13.11 \, g/cm^3 \]

### Final Answer
The density of mercury at 473 K is approximately \(13.11 \, g/cm^3\).

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