`A` spherical ball of radius `7cm` contains `56%` iron. If density is `1.4 g//cm^(3)`, number of mol of `Fe` present approximately is :

Options

A. `10`
B. `15`
C. `20`
D. `25`

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Calculate the Volume of the Spherical Ball
The formula for the volume \( V \) of a sphere is given by:

\[ V = \frac{4}{3} \pi r^3 \]

Where \( r \) is the radius of the sphere. Given that the radius \( r = 7 \, \text{cm} \), we can substitute this value into the formula.

\[ V = \frac{4}{3} \pi (7)^3 \]

Calculating \( 7^3 \):

\[ 7^3 = 343 \]

Now substituting this back into the volume formula:

\[ V = \frac{4}{3} \times \frac{22}{7} \times 343 \]

Calculating this gives:

\[ V = \frac{4 \times 22 \times 343}{3 \times 7} = \frac{30272}{21} \approx 1441.05 \, \text{cm}^3 \]

### Step 2: Calculate the Mass of the Spherical Ball
Using the density \( \rho = 1.4 \, \text{g/cm}^3 \), we can calculate the mass \( m \) of the ball using the formula:

\[ m = V \times \rho \]

Substituting the values we have:

\[ m = 1441.05 \, \text{cm}^3 \times 1.4 \, \text{g/cm}^3 \]

Calculating this gives:

\[ m \approx 2017.47 \, \text{g} \]

### Step 3: Calculate the Mass of Iron in the Ball
Since the ball contains \( 56\% \) iron, we calculate the mass of iron \( m_{Fe} \):

\[ m_{Fe} = \frac{56}{100} \times m \]

Substituting the mass we calculated:

\[ m_{Fe} = 0.56 \times 2017.47 \approx 1130.79 \, \text{g} \]

### Step 4: Calculate the Number of Moles of Iron
The number of moles \( n \) of iron can be calculated using the formula:

\[ n = \frac{m_{Fe}}{M_{Fe}} \]

Where \( M_{Fe} \) is the molar mass of iron, which is approximately \( 56 \, \text{g/mol} \).

Substituting the values:

\[ n = \frac{1130.79 \, \text{g}}{56 \, \text{g/mol}} \approx 20.18 \, \text{mol} \]

### Final Answer
The number of moles of iron present in the spherical ball is approximately \( 20 \, \text{mol} \).

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