What is the maximum amount of water (in grams) that can be removed from 15ml of toluene by the addition?
1 answer:
Complete Question
Magnesium sulfate forms a hydrate with the formula . What is the maximum amount of water (in grams) that can be removed from 15 ml of toluene by the addition of 200 mg of anhydrous magnesium sulfate? The molar mass of
is 120.4 g/mol; H20 = 18 g/mol.
Answer:
The value is of
Explanation:
From the question we are told that
The volume of toluene is
The mass of anhydrous magnesium sulfate is
The formula of the hydrate is
The molar mass of is
From the formula given we see that
1 mole of wil remove 7 moles of
to for the given formula
Hence
120.4 g (1 mole) will remove 7 moles (7 * 18 g = 126 g ) of to for the given formula
Therefore 1 g of x g of
So
=>
From our calculation we obtained that
1 g of will remove
of
Then
of
will remove z g of
of
So
=>
=> of
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Answer:
7.21 × 10⁴ J
Explanation:
Ethanol is solid below -114.5°c, liquid between -114.5°C and 78.4°C, and gaseous above 78.4°C.
<em>How much heat energy is required to convert 48.3 g of solid ethanol at -114.5°C to gaseous ethanol at 135.3 °C?</em>
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We need to calculate the heat required in different stages and then add them.
The moles of ethanol are:
Solid-liquid transition
Q₁ = ΔHfus . n = (4.60 kJ/mol) . 1.05 mol = 4.83 kJ = 4.83 × 10³ J
where,
ΔHfus: molar heat of fusion
n: moles
Liquid: from -114.5°C to 78.4°C
Q₂ = c(l) . m . ΔT = (2.45 J/g.°C) . 48.3g . [78.4°C-(-114.5°C)] = 2.28 × 10⁴ J
where,
c(l): specific heat capacity of the liquid
ΔT: change in the temperature
Liquid-gas transition
Q₃ = ΔHvap . n = (38.56 kJ/mol) . 1.05 mol = 40.5 kJ = 40.5 × 10³ J
where,
ΔHvap: molar heat of vaporization
Gas: from 78.4°C to 135.3°C
Q₄ = c(g) . m . ΔT = (1.43 J/g.°C) . 48.3g . (135.3°C-78.4°C) = 3.93 × 10³ J
where
c(g): specific heat capacity of the gas
Total heat required
Q₁ + Q₂ + Q₃ + Q₄ = 4.83 × 10³ J + 2.28 × 10⁴ J + 40.5 × 10³ J + 3.93 × 10³ J = 7.21 × 10⁴ J