To find the value of \( x \) in the hydrate \( CuSO_4 \cdot xH_2O \), we start with the initial mass of the hydrated salt and the mass of the anhydrous salt remaining after heating.
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Initial mass of the hydrate: \[ \text{Mass of } CuSO_4 \cdot xH_2O = 0.499 , \text{g} \]
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Mass of the anhydrous CuSO4 after heating: \[ \text{Mass of } CuSO_4 = 0.346 , \text{g} \]
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Mass of water lost during heating: \[ \text{Mass of water lost} = \text{Mass of hydrate} - \text{Mass of anhydrous salt} \] \[ = 0.499 , \text{g} - 0.346 , \text{g} = 0.153 , \text{g} \]
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Molar mass of \( CuSO_4 \):
- Copper (Cu): \( 63.55 , \text{g/mol} \)
- Sulfur (S): \( 32.07 , \text{g/mol} \)
- Oxygen (O): \( 16.00 \times 4 = 64.00 , \text{g/mol} \) \[ \text{Molar mass of } CuSO_4 = 63.55 + 32.07 + 64.00 = 159.62 , \text{g/mol} \]
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Molar mass of water \( H_2O \): \[ \text{Molar mass of } H_2O = 2 \times 1.01 + 16.00 = 18.02 , \text{g/mol} \]
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Calculate the number of moles of \( CuSO_4 \): \[ \text{Moles of } CuSO_4 = \frac{\text{Mass of } CuSO_4}{\text{Molar mass of } CuSO_4} = \frac{0.346 , \text{g}}{159.62 , \text{g/mol}} \approx 0.00217 , \text{mol} \]
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Calculate the number of moles of water lost: \[ \text{Moles of water} = \frac{\text{Mass of water lost}}{\text{Molar mass of water}} = \frac{0.153 , \text{g}}{18.02 , \text{g/mol}} \approx 0.00849 , \text{mol} \]
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Calculate the ratio of moles of water to moles of \( CuSO_4 \): \[ \text{Ratio} = \frac{\text{Moles of water}}{\text{Moles of } CuSO_4} = \frac{0.00849 , \text{mol}}{0.00217 , \text{mol}} \approx 3.91 \]
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Conclusion: Since \( x \) must be a whole number, we round \( 3.91 \) to the nearest whole number, which is \( 4 \). Therefore, the value of \( x \) in the hydrate \( CuSO_4 \cdot xH_2O \) is approximately \( 4 \).
Thus, the value of \( x \) is: \[ \boxed{4} \]