What is the probable gas if it takes half the time to effuse as helium?
To determine the probable gas if it takes half the time to effuse as helium, we can use Graham's law of effusion. According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
In this case, we know that the gas in question takes half the time to effuse compared to helium. Since helium is a known gas with a molar mass of approximately 4 grams per mole, we can set up the following equation:
Rate of the unknown gas / Rate of helium = √(Molar mass of helium / Molar mass of the unknown gas)
Given that the rate of the unknown gas is half the rate of helium, we can express it as:
1/2 = √(4 / Molar mass of the unknown gas)
Squaring both sides of the equation, we get:
1/4 = 4 / Molar mass of the unknown gas
Cross-multiplying and rearranging the equation, we find:
Molar mass of the unknown gas = 16
Therefore, the probable gas in question has a molar mass of 16 grams per mole. By consulting a periodic table, we can identify that this molar mass corresponds to the gas sulfur dioxide (SO2).