0.2g of a gas ‘x’ occupy a volume of 440ml. If 0.1g of CO2 gas occupy a volume of 320ml at the same temperature and pressure, then the gas ‘x’ could be

So, we know that, PV = nRTNow, V = nRT/P If. P and T are constant then,V2/V1=n2/n1or, putting the values we get,0.32/0.44=0.002272/n1n1 =0.002272x 0.440/32=0.003124Since, 0.003124 moles of X contains = 0.2 gSo, Molar mass of the gas = given mass/ no. of moles = 0.2/0.003124= 64 gSince, Molar mass of SO2 = 32+ 16 x 2 = 64 g

0.2g of a gas ‘x’ occupy a volume of 440ml. If 0.1g of CO2 gas occupy a volume of 320ml at the same temperature and pressure, then the gas ‘x’ could be

So, we know that, PV = nRTNow, V = nRT/P If. P and T are constant then,V2/V1=n2/n1or, putting the values we get,0.32/0.44=0.002272/n1n1 =0.002272x 0.440/32=0.003124Since, 0.003124 moles of X contains = 0.2 gSo, Molar mass of the gas = given mass/ no. of moles = 0.2/0.003124= 64 gSince, Molar mass of SO2 = 32+ 16 x 2 = 64 g

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0.2g of a gas ‘x’ occupy a volume of 440ml. If 0.1g of CO₂ gas occupy a volume of 320ml at the same temperature and pressure, then the gas ‘x’ could be

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SO₂

O₂

C₄H₁₀

NO₂

answer is D.

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Detailed Solution

So, we know that, PV = nRT
Now, V = nRT/P
If. P and T are constant then,
V₂/V₁=n₂/n₁
or, putting the values we get,0.32/0.44=0.002272/n₁
n₁ =0.002272x 0.440/32=0.003124
Since, 0.003124 moles of X contains = 0.2 g
So, Molar mass of the gas = given mass/ no. of moles = 0.2/0.003124= 64 g
Since, Molar mass of SO₂ = 32+ 16 x 2 = 64 g