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If the sample contains 38% of Cl2.Then, the mass of Cl2 present in sample = 38 % of 50 gm = 19 gmMolar mass of Cl2 = 71 gNumber of moles of Cl2 = 71/19 = 3.74At STP, 1 mole Cl2 = 22.4 LSo, Volume of gas = 22.4 x 3.74 = 83.8 L
If the sample contains 38% of Cl2.
Then, the mass of Cl2 present in sample = 38 % of 50 gm = 19 gm
Molar mass of Cl2 = 71 g
Number of moles of Cl2 = 71/19 = 3.74
At STP, 1 mole Cl2 = 22.4 L
So, Volume of gas = 22.4 x 3.74 = 83.8 L
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