`2.3xx10^(22)` atoms of an element weigh 6.9 g. Atomic weight of that element is

Options

A. 290 g
B. 180 g
C. 34.4 g
D. 10.4 g

Explanation

To find the atomic weight of the element given that \(2.3 \times 10^{22}\) atoms weigh 6.9 g, we can follow these steps:

### Step 1: Understand the relationship between moles, atoms, and atomic weight
The atomic weight of an element is defined as the mass of one mole of atoms of that element. One mole of any substance contains \(6.022 \times 10^{23}\) entities (atoms, molecules, etc.), known as Avogadro's number.

### Step 2: Calculate the mass of one mole of atoms
We know that:
- \(2.3 \times 10^{22}\) atoms weigh 6.9 g.
- To find the weight of \(6.022 \times 10^{23}\) atoms (1 mole), we can set up a proportion.

Using the formula:
\[ \text{Weight of 1 mole} = \left(\frac{\text{Weight of given atoms} \times \text{Avogadro's number}}{\text{Number of given atoms}}\right) \]

Substituting the values:
\[ \text{Weight of 1 mole} = \left(\frac{6.9 \, \text{g} \times 6.022 \times 10^{23}}{2.3 \times 10^{22}}\right) \]

### Step 3: Perform the calculation
Now, we can calculate the above expression step by step.

1. Calculate the numerator:
\[ 6.9 \, \text{g} \times 6.022 \times 10^{23} = 41.54 \times 10^{23} \, \text{g} \]

2. Now divide by the number of atoms:
\[ \frac{41.54 \times 10^{23} \, \text{g}}{2.3 \times 10^{22}} = 180 \, \text{g} \]

### Step 4: Conclusion
Thus, the atomic weight of the element is \(180 \, \text{g/mol}\).

### Final Answer:
The atomic weight of the element is \(180 \, \text{g/mol}\).

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