High School

A 39.4 g sample of halothane is 0.200 moles. At STP, what is the concentration (% v/v) of 39.4 g of halothane that has been completely vaporized into a container with a volume of 224 L? Assume that halothane is an ideal gas.

Answer :

To find the concentration (% v/v) of halothane at STP, multiply the number of moles (0.200) by the molar volume (22.4 L/mol) to find the volume of halothane gas (4.48 L). Divide this volume by the container's volume (224 L) and multiply by 100 to get a concentration of 2% by volume.

To calculate the concentration (% v/v) of halothane gas at STP, we first need to determine its volume. Given that we have 0.200 moles of halothane and one mole of an ideal gas occupies 22.4 liters at STP, we can calculate the volume of halothane:

[tex]Volume\ of\ Halothane = 0.200\ moles \times 22.4\ L/mole = 4.48\ liters[/tex]

Next, we calculate the concentration (% v/v) by dividing the volume of halothane by the total volume of the container and multiplying by 100:

[tex]Concentration\ (\% v/v) = \frac{4.48\ L}{224\ L} \times 100 = 2\%[/tex]

Therefore, the concentration of halothane gas in the container at STP is 2% by volume.

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