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Exercises · 5.9

Q.Density of a gas is found to be 5.46 g/dm3 at 27 °C at 2 bar pressure. What will be its density at STP?

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Step 1 – Find the molar mass from the given conditions

d=pMRT  ⟹  M=dRTpd = \frac{pM}{RT} \implies M = \frac{dRT}{p}

d=5.46 g dm−3,T=27+273=300 K,p=2 bar,R=0.0831 bar dm3K−1mol−1d = 5.46\ \text{g dm}^{-3}, \quad T = 27+273 = 300\ \text{K}, \quad p = 2\ \text{bar}, \quad R = 0.0831\ \text{bar dm}^3\text{K}^{-1}\text{mol}^{-1}

M=(5.46)(0.0831)(300)2=136.122=68.06 g mol−1M = \frac{(5.46)(0.0831)(300)}{2} = \frac{136.12}{2} = 68.06\ \text{g mol}^{-1}

Step 2 – Use this molar mass (a fixed property of the gas) to find density at STP …

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