Iodine, i2, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. what is the temperature (in k) in a 71.4 ml bulb that contains 0.276 g of i2 vapor at a pressure of 0.478 atm?

382.63 K

We are given;

We are required to calculate the temperature of Iodine

T = PV ÷ nR

But, n, the number of moles = Mass ÷ Molar mass

Molar mass of iodine = 253.8089 g/mol

Thus, n = 0.276 g ÷ 253.8089 g/mol

           = 0.001087 moles

Therefore;

T = (0.478 atm × 0.0714 L) ÷ (0.001087 moles × 0.082057)

  = 382.63 K

Thus, the temperature of Iodine in Kelvin is 382.63 K

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