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svetlana [45]
3 years ago
14

A student makes a compound of sulfur and oxygen. She uses 5.00 g of sulfur and 4.99 g of oxygen, and all of the elemental substa

nces are completely used up. What is the percent sulfur in the compound?
Chemistry
1 answer:
victus00 [196]3 years ago
7 0
First, we apply the law of conservation of mass which states that the total mass in a system remains constant.
Therefore, there must be 5.00 g of sulfur and 4.99 g of oxygen in the product. Now, we determine the mass percentage using:

Mass % = (mass of sulfur x 100) / total mass of compound

Mass % = (5 * 100) / (5 + 4.99)

Mass % = 50.05%

The product contains 50.05% sulfur by mass.
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Answer : The vapor pressure (in atm) of a solution is, 0.679 atm

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Mass of H_2O = 1.00 kg = 1000 g

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First we have to calculate the moles of H_2O.

\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{1000g}{18g/mole}=55.55mole

Now we have to calculate the mole fraction of H_2O

\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CsF}=\frac{55.55}{55.55+3.68}=0.938

Now we have to partial pressure of solution.

According to the Raoult's law,

P_{Solution}=X_{H_2O}\times P^o_{H_2O}

where,

P_{Solution} = vapor pressure of solution

P^o_{H_2O} = vapor pressure of water = 0.692 atm

X_{H_2O} = mole fraction of water = 0.938

P_{Solution}=X_{H_2O}\times P^o_{H_2O}

P_{Solution}=0.938\times 0.692atm

P_{Solution}=0.649atm

Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm

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