Acetylene, C2H2, burns according to the following reaction: C2H2 5O2 --> 4CO2 2H2O. Suppose 1.20 g of C2H2 is mixed with 3.50
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The complete question is as follows: 171 g of sucrose ( MW of 342, melting point 186 oC, boiling point very high, and vapor pressure is negligible) is dissolved in one liter of water at 25 oC. At 25 oC the vapor pressure of water is 24 mmHg. Which value is closest to the vapor pressure (VP) of this solution at 25 oC?
a. 16mm Hg
b. 24mm Hg
c. 20mm Hg
d. 12mm Hg
Answer: The vapor pressure (VP) of this solution at is closest to the value 24 mm Hg.
Explanation:
Given: Mass of sucrose = 171 g
Mass of water = 1 L = 1000 g
Vapor pressure of water = 24 mm Hg
As moles is the mass of substance divided by its molar mass. Hence, moles of water (molar mass = 18.02 g) is calculated as follows.
Similarly, moles of sucrose (molar mass = 342 g/mol) is as follows.
Total moles = 55.49 + 0.5 mol = 55.99 mol
Mole fraction of water is as follows.
Formula used to calculate vapor pressure of the solution is as follows.
where,
= vapor pressure of component i over the solution
= vapor pressure of pure component i
= mole fraction of i
Substitute the values into above formula to calculate vapor pressure of water as follows.
Thus, we can conclude that the vapor pressure (VP) of this solution at is closest to the value 24 mm Hg.
Answer : The molar mass of the unknown gas will be 79.7 g/mol
Explanation : To solve this question we can use graham's law;
Now we can use nitrogen as the gas number 2, which travels faster than gas 1;
So, 167 / 99 = 1.687 So the nitrogen gas is 1.687 times faster that the unknown gas 1
We can compare the rates of both the gases;
So here, Rate of gas 2 / Rate of gas 1 =
Now, 1.687 = square root [ ]
When we square both the sides we get;
2.845 = (molar mass 1) / (28.01 g/mol N2)
On rearranging, we get,
2.845 X (28.01 g/mol N2) = Molar mass 1
So the molar mass of unknown gas will be = 79.7 g/mol