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## anonymous one year ago At STP, 0.106 g of a gas sample occupies a volume of 62.2 mL. What is the molar mass of this gas? Keep 3 sig figs for your final answer.

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1. anonymous

Is it three sig figs or four?

2. Photon336

Start with the ideal gas equation; we need number of moles n so we solve for n $pV = nRT$ then our equation becomes this $\frac{ pV }{ RT } = n$ $STP = 1 atm, 298K, 25C = Standard temperature and pressure$ plug everything in $\frac{ pV }{ RT } = n \frac{ 1atm*0.062L }{ (0.08L, atm, mol, * 298) } = 2.54x10^{-3} mol$ we know that 0.00254 moles contains .106 grams but how many grams are in 1 mole. we can do this by setting up a proportion. $\frac{ mol }{ x,grams } = \frac{ 0.00254,moles }{ 0.106 grams }$ cross multiply $0.00254 moles, x = 1.06 grams$ 1 mol contains 44.732 grams $\frac{ 0.106grams }{ 0.00254mol } = 44.732 grams$

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