Answer :
Answer:
The formula is Be(NO3)2*3H2O. The number of waters in the hydrate is 3
Explanation:
Step 1: Data given
Mass of the hydrate sample = 3.41 grams
Mass after heating = 2.43 grams
Step 2: Calculate mass of water
After heating, all the water is gone. So the mass of water can be calculated by
Mass of hydrate before heating - mass after heating
Mass of water = 3.41 -2.43 = 0.98 grams
Step 2 : Calculate moles H2O
Moles H2O = mass H2O / molar mass H2O
Moles H2O = 0.98 grams / 18.02 g/mol
Moles H2O = 0.054 moles
Step 3: Calculate moles Be(NO3)2
Moles Be(NO3)2 = 2.43 grams / 133.02
Moles Be(NO3)2 = 0.0183 moles
Step 4: Calculate molecules water
Molecules H2O = 0.054 moles / 0.0183 moles
Molecules H2O = 3.0
The formula is Be(NO3)2*3H2O. The number of waters in the hydrate is 3
The number of waters of hydration in the hydrate is 3.
Based on the given information,
• The formula of the beryllium nitrate hydrate is Be(NO3)2⋅xH2O.
• The weight of the sample is 3.41 grams.
• The weight of the sample after heating of the sample is 2.43 grams {weight of Be(NO3)2}.
Now, the mass of xH2O will be,
= 3.41 g - 2.43 g
= 0.98 grams
The moles of Be(NO3)2 will be calculated as,
Mass of Be(NO3)2/Molecular weight of Be(NO3)2 = 2.43 g/133 = 0.01827 moles
The moles of xH2O = 0.98/18 = 0.05444 moles
Now the value of x will be,
= 0.05444/0.01827
= 3
Thus, the number of waters of hydration in the hydrate is 3.
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