1.00 g of a mixture, consisting of equal number of moles of carbonates of two alkali metals, required 44.4 " mL of " 0.5 N-HCl for complete reaction. If the atomic weight of one of the metal is 7.00. Find the atomic weight of the other metal.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the Variables
Let the two alkali metals be M1 and M2. The atomic weight of M1 is given as 7.00 g/mol. The carbonates of these metals can be represented as:
- M1CO3 (for the first metal)
- M2CO3 (for the second metal)

Let the atomic weight of M2 be \( m \).
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