`100 g` sample of clay (containing 19% `H_(2)O`, 40% silica, and inert inpurities as rest) is partically dried so as to contains 10% `H_(2) O`.
Which of the following `is//are` correct statements (s) ?

Options

A. The percentage of silica in it is 44.4%
B. The mas of partically dried clay is `90.0 g`.
C. The precentage of inert impurity in it is 45.6%
D. The mass of water evaporated is `10.0 g`

Explanation

To solve the problem step by step, we need to analyze the given information about the clay sample and its composition before and after drying.

### Step 1: Determine the Composition of the Original Clay Sample
The original clay sample weighs 100 g and contains:
- 19% H₂O
- 40% Silica
- The rest is inert impurities.

Calculating the amounts:
- Amount of H₂O = 19% of 100 g = 19 g
- Amount of Silica = 40% of 100 g = 40 g
- Amount of Inert Impurities = 100 g - (19 g + 40 g) = 100 g - 59 g = 41 g

### Step 2: Determine the Composition of the Partially Dried Clay Sample
After partial drying, the clay sample contains 10% H₂O. We need to find the new total mass of the sample that corresponds to this percentage.

Let the mass of the partially dried clay sample be \( x \) grams. Since it contains 10% H₂O:
- Amount of H₂O in the dried sample = 10% of \( x \) = 0.1x
- The remaining mass (non-H₂O) = \( x - 0.1x = 0.9x \)

### Step 3: Calculate the Non-H₂O Components in the Dried Sample
From the original sample, we know:
- Amount of Silica = 40 g
- Amount of Inert Impurities = 41 g

The total non-H₂O mass in the original sample = 100 g - 19 g = 81 g.

Since the proportions of non-H₂O components remain the same, we can set up a proportion:
- Mass of Silica in dried sample = (40 g / 81 g) × (0.9x)
- Mass of Inert Impurities in dried sample = (41 g / 81 g) × (0.9x)

### Step 4: Find the Total Mass of the Dried Sample
We know that the total mass of the dried sample must equal the sum of the masses of H₂O, Silica, and Inert Impurities:
- \( x = 0.1x + \text{Mass of Silica} + \text{Mass of Inert Impurities} \)

Substituting the values we calculated:
- \( x = 0.1x + \left(\frac{40}{81} \cdot 0.9x\right) + \left(\frac{41}{81} \cdot 0.9x\right) \)

### Step 5: Solve for \( x \)
Combine the terms:
- \( x = 0.1x + 0.9x \left(\frac{40 + 41}{81}\right) \)
- \( x = 0.1x + 0.9x \left(\frac{81}{81}\right) \)
- \( x = 0.1x + 0.9x \)
- \( x = x \) (which confirms our setup)

### Step 6: Calculate the Mass of Water Evaporated
The mass of water evaporated = Original mass of H₂O - Mass of H₂O in dried sample
- Mass of water evaporated = 19 g - 10 g = 9 g

### Step 7: Final Results
From our calculations, we can summarize:
- Mass of Silica in dried sample = 44.4 g
- Mass of Inert Impurities in dried sample = 45.6 g
- Mass of water evaporated = 9 g

### Conclusion
The correct statements are:
1. The percentage of Silica is 44.4%.
2. The percentage of Inert Impurities is 45.6%.
3. The mass of water evaporated is 9 g.