A sample of Cl2 gas having a volume of 0.75 L at 20∘C and 725 mmHg has a mass equal to

pV=nRT=mMRTHence,m=pVMRT=725mmHg760mmHg×1atm(0.75L)71gmol−10.082atmLK−1mol−1(293K)=2.11g

A sample of Cl2 gas having a volume of 0.75 L at 20∘C and 725 mmHg has a mass equal to

pV=nRT=mMRTHence,m=pVMRT=725mmHg760mmHg×1atm(0.75L)71gmol−10.082atmLK−1mol−1(293K)=2.11g

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A sample of ${Cl}_{2}$ gas having a volume of 0.75 L at $20^{\circ} C$ and 725 mmHg has a mass equal to

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1600 g

10.6 g

2.11 g

5.3 g

answer is C.

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Detailed Solution

$p V = n R T = \frac{m}{M} R T$

Hence,

$m = \frac{p V M}{R T} = \frac{(\frac{725 mmHg}{760 mmHg} \times 1 atm) (0.75 L) (71 g {mol}^{- 1})}{(0.082 {atmLK}^{- 1} {mol}^{- 1}) (293 K)} = 2.11 g$

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A sample of Cl2 gas having a volume of 0.75 L at 20∘C and 725 mmHg has a mass equal to

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A sample of ${Cl}_{2}$ gas having a volume of 0.75 L at $20^{\circ} C$ and 725 mmHg has a mass equal to