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Answer :
To solve the problem regarding the dehydration of the hydrate, we are going to break down each part step-by-step.
### 1. Determine the Molar Mass of the Hydrate [tex]\( \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} \)[/tex]
To find the molar mass, we need to add up the atomic masses of all the atoms in the compound. Let's calculate:
- Sodium ([tex]\(\text{Na}\)[/tex]): There are 2 sodium atoms.
- Molar mass = 2 × 22.99 g/mol = 45.98 g/mol
- Carbon ([tex]\(\text{C}\)[/tex]): There is 1 carbon atom.
- Molar mass = 12.01 g/mol
- Oxygen ([tex]\(\text{O}\)[/tex]) in [tex]\(\text{Na}_2\text{CO}_3\)[/tex]: There are 3 oxygen atoms.
- Molar mass = 3 × 16.00 g/mol = 48.00 g/mol
- Water ([tex]\(\text{H}_2\text{O}\)[/tex]): There are 10 water molecules, each containing 2 hydrogen and 1 oxygen atom.
- Molar mass of one water molecule = 2 × 1.008 g/mol (H) + 16.00 g/mol (O) = 18.016 g/mol
- Total for 10 water molecules = 10 × 18.016 g/mol = 180.16 g/mol
Total molar mass of [tex]\(\text{Na}_2\text{CO}_3 \cdot 10 \text{H}_2\text{O}\)[/tex]:
[tex]\[
45.98 \, (\text{Na}) + 12.01 \, (\text{C}) + 48.00 \, (\text{O in} \, \text{Na}_2\text{CO}_3) + 180.16 \, (\text{H}_2\text{O}) = 286.15 \, \text{g/mol}
\][/tex]
### 2. Theoretical Water Loss in Grams
Next, we determine how much water would be lost when the hydrate is dehydrated.
1. From the molar mass calculation above, we found that the water in the hydrate weighs 180.16 g/mol.
2. Given: the mass of the hydrate being dehydrated is 37.9 grams.
Using proportion to find the expected water loss:
[tex]\[
\text{Water loss} = \left( \frac{\text{Water mass in 1 mol}}{\text{Total molar mass of hydrate}} \right) \times \text{Total mass of hydrate}
\][/tex]
[tex]\[
\text{Water loss} = \left( \frac{180.16}{286.15} \right) \times 37.9 = 23.86 \, \text{grams (approx)}
\][/tex]
Thus, the students should theoretically expect to see a water loss of approximately 23.86 grams when dehydrating 37.9 grams of the hydrate [tex]\(\text{Na}_2\text{CO}_3 \cdot 10 \text{H}_2\text{O}\)[/tex].
### 1. Determine the Molar Mass of the Hydrate [tex]\( \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} \)[/tex]
To find the molar mass, we need to add up the atomic masses of all the atoms in the compound. Let's calculate:
- Sodium ([tex]\(\text{Na}\)[/tex]): There are 2 sodium atoms.
- Molar mass = 2 × 22.99 g/mol = 45.98 g/mol
- Carbon ([tex]\(\text{C}\)[/tex]): There is 1 carbon atom.
- Molar mass = 12.01 g/mol
- Oxygen ([tex]\(\text{O}\)[/tex]) in [tex]\(\text{Na}_2\text{CO}_3\)[/tex]: There are 3 oxygen atoms.
- Molar mass = 3 × 16.00 g/mol = 48.00 g/mol
- Water ([tex]\(\text{H}_2\text{O}\)[/tex]): There are 10 water molecules, each containing 2 hydrogen and 1 oxygen atom.
- Molar mass of one water molecule = 2 × 1.008 g/mol (H) + 16.00 g/mol (O) = 18.016 g/mol
- Total for 10 water molecules = 10 × 18.016 g/mol = 180.16 g/mol
Total molar mass of [tex]\(\text{Na}_2\text{CO}_3 \cdot 10 \text{H}_2\text{O}\)[/tex]:
[tex]\[
45.98 \, (\text{Na}) + 12.01 \, (\text{C}) + 48.00 \, (\text{O in} \, \text{Na}_2\text{CO}_3) + 180.16 \, (\text{H}_2\text{O}) = 286.15 \, \text{g/mol}
\][/tex]
### 2. Theoretical Water Loss in Grams
Next, we determine how much water would be lost when the hydrate is dehydrated.
1. From the molar mass calculation above, we found that the water in the hydrate weighs 180.16 g/mol.
2. Given: the mass of the hydrate being dehydrated is 37.9 grams.
Using proportion to find the expected water loss:
[tex]\[
\text{Water loss} = \left( \frac{\text{Water mass in 1 mol}}{\text{Total molar mass of hydrate}} \right) \times \text{Total mass of hydrate}
\][/tex]
[tex]\[
\text{Water loss} = \left( \frac{180.16}{286.15} \right) \times 37.9 = 23.86 \, \text{grams (approx)}
\][/tex]
Thus, the students should theoretically expect to see a water loss of approximately 23.86 grams when dehydrating 37.9 grams of the hydrate [tex]\(\text{Na}_2\text{CO}_3 \cdot 10 \text{H}_2\text{O}\)[/tex].
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