Answer:
7
Explanation:
Let x represent the number of moles of water in the hydrated salt i.e MgSO₄.xH₂O
The following data were obtained from the question:
Mass of MgSO₄.xH₂O = 1.547 g
Mass of anhydrous MgSO₄ = 0.7554 g
Mole of H₂O = x =?
Next, we shall determine the mass of water, H₂O in the hydrated salt, MgSO₄.xH₂O. This can be obtained as follow:
Mass of MgSO₄.xH₂O = 1.547 g
Mass of anhydrous MgSO₄ = 0.7554 g
Mass of H₂O =?
Mass of H₂O = (Mass of MgSO₄.xH₂O) – (Mass of anhydrous MgSO₄)
Mass of H₂O = 1.547 – 0.7554
Mass of H₂O = 0.7916 g
Finally, we shall determine the value of the x as illustrated below:
Mass of MgSO₄.xH₂O = 1.547 g
Molar mass of MgSO₄.xH₂O = 24 + 32 + (16×4) + x[(2×1) + 16]
= 24 + 32 + 64 + x(2 + 16)
= 120 + 18x
Mass of H₂O = 0.7916 g
Molar mass of xH₂O = 18x
Molar Mass of xH₂O/ Molar mass of MgSO₄.xH₂O = mass of xH₂O /Mass of MgSO₄.xH₂O
18x/ 120 + 18x = 0.7916/1.547
Cross multiply
0.7916 (120 + 18x) = 18x × 1.547
94.992 + 14.2488x = 27.846x
Collect like terms
94.992 = 27.846x – 14.2488x
94.992 = 13.5972x
Divide both side by 13.5972
x = 94.992 / 13.5972
x = 7
Thus, the formula for the hydrated salt, MgSO₄.xH₂O is MgSO₄.7H₂O
Number of moles of water, H₂O in the hydrated salt MgSO₄.7H₂O is 7.
The number of moles of attached water molecules is 7.
Mass of hydrated MgSO4 = 1.547 grams
Mass of anhydrous MgSO4 = 0.7554 grams
Number of moles of hydrated MgSO4 = 1.547 grams/120 + 18x
Number of moles of anhydrous MgSO4 = 0.7554 grams /120
Number of moles of anhydrous salt = Number of moles of hydrated salt
0.7554 grams /120 = 1.547 grams/120 + 18x
0.7554(120 + 18x) = 1.547 × 120
90.6 + 13.6x = 185.6
185.6 - 90.6 /13.6 = x
x = 7
The number of moles of attached water molecules is 7.
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