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The number of atoms of Mg, S and O prese...

The number of atoms of Mg, S and O present in 0.5 moles of `MgSO_4` are respectively
(Given : Atomic mass of O= 16 u , Mg = 24 u , S= 32 u)

A

`3.01 xx 10^(23) , 3.01 xx 10^(23) and 12.04 xx 10^(23)`

B

`3xx10^(23) , 3 xx 10^(23) and 1.2 xx 10^(23)`

C

`3.01xx10^(23) , 12.05 xx 10^(23) and 12.04 xx 10^(23)`

D

`3xx 10^(23) , 12.05 xx 10^(23) and 12.04 xx 10^(23)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the number of atoms of magnesium (Mg), sulfur (S), and oxygen (O) present in 0.5 moles of magnesium sulfate (MgSO₄), we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Composition of MgSO₄**: - The formula MgSO₄ indicates that each molecule of magnesium sulfate contains: - 1 atom of Magnesium (Mg) - 1 atom of Sulfur (S) - 4 atoms of Oxygen (O) 2. **Calculate the Number of Molecules in 0.5 Moles**: - Using Avogadro's number, which is approximately \(6.02 \times 10^{23}\) molecules/mole, we can calculate the number of molecules in 0.5 moles: \[ \text{Number of molecules} = 0.5 \text{ moles} \times 6.02 \times 10^{23} \text{ molecules/mole} = 3.01 \times 10^{23} \text{ molecules} \] 3. **Calculate the Number of Atoms for Each Element**: - **For Magnesium (Mg)**: - Since there is 1 atom of Mg per molecule: \[ \text{Number of Mg atoms} = 3.01 \times 10^{23} \text{ molecules} \times 1 = 3.01 \times 10^{23} \text{ atoms} \] - **For Sulfur (S)**: - Since there is 1 atom of S per molecule: \[ \text{Number of S atoms} = 3.01 \times 10^{23} \text{ molecules} \times 1 = 3.01 \times 10^{23} \text{ atoms} \] - **For Oxygen (O)**: - Since there are 4 atoms of O per molecule: \[ \text{Number of O atoms} = 3.01 \times 10^{23} \text{ molecules} \times 4 = 1.204 \times 10^{24} \text{ atoms} \] 4. **Summarize the Results**: - The number of atoms of Mg, S, and O in 0.5 moles of MgSO₄ are: - Mg: \(3.01 \times 10^{23}\) atoms - S: \(3.01 \times 10^{23}\) atoms - O: \(1.204 \times 10^{24}\) atoms ### Final Answer: The number of atoms of Mg, S, and O present in 0.5 moles of MgSO₄ are respectively: - Mg: \(3.01 \times 10^{23}\) - S: \(3.01 \times 10^{23}\) - O: \(1.204 \times 10^{24}\)
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