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. `13.12 g//c.c.`
B. `13.65 g//c.c.`
C. `13.51 g//c.c.`
D. `13.22 g//c.c.`

Explanation

To find the density of mercury at 473 K given its coefficient of apparent expansion and its density at 0 °C, we can follow these steps:

### Step 1: Understand the Given Information
- Coefficient of apparent expansion of mercury, \( \alpha = 0.18 \times 10^{-3} \, \text{°C}^{-1} \)
- Density of mercury at 0 °C, \( d_0 = 13.6 \, \text{g/cm}^3 \)
- Temperature at which we need to find the density, \( T = 473 \, \text{K} \)

### Step 2: Convert Temperature to Celsius
To convert the temperature from Kelvin to Celsius:
\[ T(°C) = T(K) - 273 \]
Thus,
\[ T(°C) = 473 - 273 = 200 \, \text{°C} \]

### Step 3: Calculate the Change in Temperature
Now, we find the change in temperature \( \Delta T \):
\[ \Delta T = T(°C) - 0 = 200 \, \text{°C} \]

### Step 4: Use the Density Formula
The formula for the density of a substance considering its apparent expansion is given by:
\[ d = \frac{d_0}{1 + \alpha \Delta T} \]
Substituting the known values:
\[ d = \frac{13.6 \, \text{g/cm}^3}{1 + (0.18 \times 10^{-3}) \times 200} \]

### Step 5: Calculate the Denominator
First, calculate \( \alpha \Delta T \):
\[ \alpha \Delta T = 0.18 \times 10^{-3} \times 200 = 0.036 \]
Now, calculate the denominator:
\[ 1 + 0.036 = 1.036 \]

### Step 6: Calculate the Density
Now substitute back into the density formula:
\[ d = \frac{13.6}{1.036} \]
Calculating this gives:
\[ d \approx 13.12 \, \text{g/cm}^3 \]

### Step 7: Final Result
Thus, the density of mercury at 473 K is approximately:
\[ d \approx 12.127 \, \text{g/cm}^3 \]