The volume of neon gas in cm at STP having the same number of atoms as that present in 800 mg of Ca is (At. mass : `Ca = 40 g mol^(-1), Ne = 20 g mol^(-1)`)

Options

A. 56
B. 896
C. 224
D. 448

Explanation

To solve the problem of finding the volume of neon gas at STP that has the same number of atoms as present in 800 mg of calcium, we can follow these steps:

### Step 1: Convert the mass of calcium from mg to grams.
Given mass of calcium = 800 mg

To convert mg to grams:
\[ 800 \text{ mg} = 800 \times 10^{-3} \text{ g} = 0.800 \text{ g} \]

### Step 2: Calculate the number of moles of calcium.
The molar mass of calcium (Ca) is given as 40 g/mol.

Using the formula for moles:
\[ \text{Number of moles of Ca} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.800 \text{ g}}{40 \text{ g/mol}} = 0.02 \text{ mol} \]

### Step 3: Calculate the number of atoms in calcium.
Using Avogadro's number (\(N_A = 6.022 \times 10^{23} \text{ atoms/mol}\)):
\[ \text{Number of atoms in Ca} = \text{Number of moles} \times N_A = 0.02 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} = 1.2044 \times 10^{22} \text{ atoms} \]

### Step 4: Determine the volume of neon gas at STP.
At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters.

To find the volume of gas that has the same number of atoms:
1 mole of neon (Ne) has \(N_A\) atoms, so:
\[ \text{Volume of 1 atom of Ne} = \frac{22.4 \text{ L}}{N_A} \]

Thus, the volume for \(0.02\) moles of neon gas (which has the same number of atoms as the calcium):
\[ \text{Volume of Ne} = 0.02 \text{ mol} \times 22.4 \text{ L/mol} = 0.448 \text{ L} \]

### Step 5: Convert the volume from liters to cm³.
Since \(1 \text{ L} = 1000 \text{ mL} = 1000 \text{ cm}^3\):
\[ 0.448 \text{ L} = 0.448 \times 1000 \text{ cm}^3 = 448 \text{ cm}^3 \]

### Final Answer:
The volume of neon gas at STP having the same number of atoms as that present in 800 mg of calcium is **448 cm³**.

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