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An unknown liquid is vaporized in a 375-mL flask by immersion in a water bath, whose temperature is 99 °C, its barometric pressure is 753 torr, and its mass is 1.362 g, has a molar mass of 112 g/mol.
Once the liquid is vaporized, we can assume it behaves as an ideal gas. First, we will convert 99 °C to Kelvin using the following expression.
[tex]K = \° C + 273.15 = 99 \° C + 273.15 = 372 K[/tex]
Then, we will convert 375 mL to L using the conversion factor 1 L = 1000 mL.
[tex]375 mL \times \frac{1L}{1000mL} = 0.375 L[/tex]
Next, we can calculate the moles of the gas (n) using the ideal gas equation.
[tex]P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T} = \frac{753torr \times 0.375 L}{(62.4 torr.L/mol.K) \times 372 K} = 0.0122 mol[/tex]
0.0122 moles of the gas have a mass of 1.362 g. The molar mass (M) of the gas is:
[tex]M = \frac{m}{n} = \frac{1.362g}{0.0122mol} = 112 g/mol[/tex]
An unknown liquid is vaporized in a 375-mL flask by immersion in a water bath, whose temperature is 99 °C, its barometric pressure is 753 torr, and its mass is 1.362 g, has a molar mass of 112 g/mol.
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