What is the volume occupied by 0.155 g of boron trifluoride gas,BF3,at STP?

Name: What is the volume occupied by 0.155 g of boron trifluoride gas,BF3,at STP?
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Question

Quizplus · Accepted Answer

The molar mass of BF3 is 67.81 g/mol. Therefore, 0.155 g of BF3 is equal to 0.155 g / 67.81 g/mol = 0.00228 mol. At STP, 1 mol of any gas occupies 22.4 L. Therefore, 0.00228 mol of BF3 will occupy 0.00228 mol * 22.4 L/mol = 0.0512 L = 51.2 mL.

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