2. A hydrate, 0.700 g, was heated to drive off water. The anhydrous salt remaining weighed 0.570 g. What was the % water in the hydrate? Show all work and include correct number of significant figures.

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**Hydrate Analysis Problem**

A hydrate sample weighing 0.700 g was heated to remove the water content. After heating, the remaining anhydrous salt weighed 0.570 g. Your task is to calculate the percentage of water in the hydrate. Ensure that your answer includes all calculations with the correct number of significant figures.

**Solution Approach:**

1. **Determine the Mass of Water Lost:**

   \[
   \text{Mass of water} = \text{Initial mass of hydrate} - \text{Mass of anhydrous salt}
   \]

   \[
   \text{Mass of water} = 0.700\, \text{g} - 0.570\, \text{g} = 0.130\, \text{g}
   \]

2. **Calculate the Percentage of Water in the Hydrate:**

   \[
   \%\text{water} = \left(\frac{\text{Mass of water}}{\text{Initial mass of hydrate}}\right) \times 100\%
   \]

   \[
   \%\text{water} = \left(\frac{0.130\, \text{g}}{0.700\, \text{g}}\right) \times 100\% \approx 18.57\%
   \]

**Conclusion:**

The percent water in the hydrate is approximately 18.57%. Ensure that significant figures reflect the precision of the given data, which in this case is to three significant figures.
Transcribed Image Text:**Hydrate Analysis Problem** A hydrate sample weighing 0.700 g was heated to remove the water content. After heating, the remaining anhydrous salt weighed 0.570 g. Your task is to calculate the percentage of water in the hydrate. Ensure that your answer includes all calculations with the correct number of significant figures. **Solution Approach:** 1. **Determine the Mass of Water Lost:** \[ \text{Mass of water} = \text{Initial mass of hydrate} - \text{Mass of anhydrous salt} \] \[ \text{Mass of water} = 0.700\, \text{g} - 0.570\, \text{g} = 0.130\, \text{g} \] 2. **Calculate the Percentage of Water in the Hydrate:** \[ \%\text{water} = \left(\frac{\text{Mass of water}}{\text{Initial mass of hydrate}}\right) \times 100\% \] \[ \%\text{water} = \left(\frac{0.130\, \text{g}}{0.700\, \text{g}}\right) \times 100\% \approx 18.57\% \] **Conclusion:** The percent water in the hydrate is approximately 18.57%. Ensure that significant figures reflect the precision of the given data, which in this case is to three significant figures.
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