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A certain hydrate has the formula MgSO4.xH2O. a quantity of 54.2 g of the compound is heated in an over to drive off the water. If the steam generated exerts a pressure of 24.8 atm in a 2.0 L container at 120°C, calculate x:
- a)2
- b)5
- c)7
- d)6
Correct answer is option 'C'. Can you explain this answer?
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A certain hydrate has the formula MgSO4.xH2O. a quantity of 54.2 g of ...
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A certain hydrate has the formula MgSO4.xH2O. a quantity of 54.2 g of ...
Given:
Formula of hydrate = MgSO4.xH2O
Weight of hydrate = 54.2 g
Pressure of steam generated = 24.8 atm
Volume of container = 2.0 L
Temperature = 120°C
To find: the value of x in MgSO4.xH2O
Explanation:
When the hydrate is heated, the water is driven off and the anhydrous salt is left behind. The weight of the anhydrous salt is found by subtracting the weight of water from the weight of the hydrate.
Step 1: Calculate the weight of water in the hydrate
The formula of the hydrate shows that there are x moles of water for every mole of MgSO4. Therefore, the weight of water in the hydrate is:
18x g/mol × (54.2 g / 120.4 g/mol) = 8.10x g
Step 2: Calculate the weight of anhydrous salt left behind
The weight of the anhydrous salt is the difference between the weight of the hydrate and the weight of water driven off:
54.2 g - 8.10x g = 46.1 g
Step 3: Calculate the number of moles of anhydrous salt
The molar mass of MgSO4 is 120.4 g/mol, so the number of moles of anhydrous salt is:
46.1 g / 120.4 g/mol = 0.382 mol
Step 4: Calculate the pressure of the anhydrous salt at the given temperature and volume
The ideal gas law can be used to calculate the pressure of the anhydrous salt at the given temperature and volume:
PV = nRT
P = nRT/V
P = (0.382 mol) × (0.0821 L·atm/(mol·K)) × (393 K) / (2.0 L)
P = 7.12 atm
Step 5: Calculate the value of x
The pressure of the anhydrous salt calculated in Step 4 must be equal to the pressure of the steam generated by the water driven off in Step 1:
7.12 atm = 24.8 atm × (8.10x g / (18x g/mol) / (2.0 L / 22.4 L/mol) )
7.12 atm = 1.10x
x = 6.47
The value of x is closest to 6, so the answer is (c) 6.
Formula of hydrate = MgSO4.xH2O
Weight of hydrate = 54.2 g
Pressure of steam generated = 24.8 atm
Volume of container = 2.0 L
Temperature = 120°C
To find: the value of x in MgSO4.xH2O
Explanation:
When the hydrate is heated, the water is driven off and the anhydrous salt is left behind. The weight of the anhydrous salt is found by subtracting the weight of water from the weight of the hydrate.
Step 1: Calculate the weight of water in the hydrate
The formula of the hydrate shows that there are x moles of water for every mole of MgSO4. Therefore, the weight of water in the hydrate is:
18x g/mol × (54.2 g / 120.4 g/mol) = 8.10x g
Step 2: Calculate the weight of anhydrous salt left behind
The weight of the anhydrous salt is the difference between the weight of the hydrate and the weight of water driven off:
54.2 g - 8.10x g = 46.1 g
Step 3: Calculate the number of moles of anhydrous salt
The molar mass of MgSO4 is 120.4 g/mol, so the number of moles of anhydrous salt is:
46.1 g / 120.4 g/mol = 0.382 mol
Step 4: Calculate the pressure of the anhydrous salt at the given temperature and volume
The ideal gas law can be used to calculate the pressure of the anhydrous salt at the given temperature and volume:
PV = nRT
P = nRT/V
P = (0.382 mol) × (0.0821 L·atm/(mol·K)) × (393 K) / (2.0 L)
P = 7.12 atm
Step 5: Calculate the value of x
The pressure of the anhydrous salt calculated in Step 4 must be equal to the pressure of the steam generated by the water driven off in Step 1:
7.12 atm = 24.8 atm × (8.10x g / (18x g/mol) / (2.0 L / 22.4 L/mol) )
7.12 atm = 1.10x
x = 6.47
The value of x is closest to 6, so the answer is (c) 6.
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A certain hydrate has the formula MgSO4.xH2O. a quantity of 54.2 g of the compound is heated in an over to drive off the water. If the steam generated exerts a pressure of 24.8 atm in a 2.0 L container at 120°C, calculate x:a)2b)5c)7d)6Correct answer is option 'C'. Can you explain this answer?
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