A cup contains 250 g of water. Find the total positive charge present in the cup of water.

Options

A. `1.34xx10^(19) C`
B. `1.34xx10^(7) C`
C. `2.43xx10^(19) C`
D. `2.43xx10^(7) C`

Explanation

To find the total positive charge present in a cup containing 250 g of water, we can follow these steps:

### Step 1: Determine the composition of water
Water (H₂O) consists of two hydrogen (H) atoms and one oxygen (O) atom. Each hydrogen atom has 1 proton, and the oxygen atom has 8 protons. Therefore, the total number of protons in one molecule of water is:
- Protons from hydrogen: 2 (1 from each H)
- Protons from oxygen: 8

Total protons in one water molecule = 2 + 8 = 10 protons.

### Step 2: Calculate the charge of one proton
The charge of one proton is approximately \(1.6 \times 10^{-19}\) coulombs.

### Step 3: Calculate the total charge from protons in one water molecule
Since there are 10 protons in one water molecule, the total positive charge from one water molecule is:
\[ \text{Charge from one molecule} = 10 \times (1.6 \times 10^{-19}) = 1.6 \times 10^{-18} \text{ coulombs} \]

### Step 4: Calculate the number of moles of water in 250 g
The molar mass of water (H₂O) is:
- Hydrogen: 1 g/mol × 2 = 2 g/mol
- Oxygen: 16 g/mol
- Total = 2 + 16 = 18 g/mol

Now, we can calculate the number of moles of water in 250 g:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{250 \text{ g}}{18 \text{ g/mol}} \approx 13.89 \text{ moles} \]

### Step 5: Calculate the total number of water molecules
Using Avogadro's number (\(6.02 \times 10^{23}\) molecules/mol), we can find the total number of water molecules:
\[ \text{Total molecules} = \text{Number of moles} \times \text{Avogadro's number} = 13.89 \times (6.02 \times 10^{23}) \approx 8.36 \times 10^{24} \text{ molecules} \]

### Step 6: Calculate the total positive charge in 250 g of water
Now, we can find the total positive charge in the water:
\[ \text{Total positive charge} = \text{Total molecules} \times \text{Charge from one molecule} = 8.36 \times 10^{24} \times (1.6 \times 10^{-18}) \approx 1.34 \times 10^{7} \text{ coulombs} \]

### Final Answer
The total positive charge present in the cup of water is approximately \(1.34 \times 10^{7}\) coulombs.

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