A sample of water contains x % of `D_(2)O`. Its molecular weight is 19. The value of‘x’ is

Options

A. 25
B. 50
C. 33.33
D. 75

Explanation

To solve the problem, we need to determine the percentage of D₂O in a sample of water that has an overall molecular weight of 19.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that the sample of water contains x% of D₂O. Therefore, if we consider 1 mole of the water sample, the amount of D₂O in it will be \( \frac{x}{100} \) moles.

2. **Calculating Moles of D₂O**:
The number of moles of D₂O in the sample is given by:
\[ \text{Moles of D₂O} = \frac{x}{100} \]

3. **Calculating Moles of H₂O**:
Since the total moles of the sample is 1, the moles of H₂O will be:
\[ \text{Moles of H₂O} = 1 - \frac{x}{100} \]

4. **Molecular Weights**:
- The molecular weight of D₂O (Deuterium Oxide) is approximately 20 g/mol (2 for D and 16 for O).
- The molecular weight of H₂O (Water) is approximately 18 g/mol (2 for H and 16 for O).

5. **Setting Up the Equation**:
The total molecular weight of the mixture can be expressed as:
\[ \text{Total Molecular Weight} = (\text{Moles of D₂O} \times \text{Molecular Weight of D₂O}) + (\text{Moles of H₂O} \times \text{Molecular Weight of H₂O}) \]
Substituting the values we have:
\[ 19 = \left(\frac{x}{100} \times 20\right) + \left(1 - \frac{x}{100}\right) \times 18 \]

6. **Expanding the Equation**:
Expanding the terms gives:
\[ 19 = \frac{20x}{100} + (18 - \frac{18x}{100}) \]
Simplifying further:
\[ 19 = \frac{20x}{100} + 18 - \frac{18x}{100} \]
\[ 19 = 18 + \frac{2x}{100} \]

7. **Isolating x**:
Rearranging the equation:
\[ 19 - 18 = \frac{2x}{100} \]
\[ 1 = \frac{2x}{100} \]
Multiplying both sides by 100:
\[ 100 = 2x \]
Dividing by 2:
\[ x = 50 \]

### Conclusion:
The value of \( x \) is 50. Therefore, the sample of water contains 50% D₂O.