Question

0.2gram of a gas occupies X occupies a volume of 440ml.if 0.1 gram of CO2 occupies volume of 320 ml at same temp and pressure, gas X could be 1---SO2 2---NO 3----CO 4----NH3

Answer 1

We have, moles of CO2 (n2)= given mass/ molar mass =0.1/44 =0.002272
Now, moles of X = n1
Similarly, volume of X (V1) = 440ml =0.44L 
 volume of X (V2) =320ml =0.32L 
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 x2 = 64 g
Hence. the correct answer is 1. (SO2)

Answer 2

Answer: SO2

Solution: Direct Mathod For Competitions

V1/V2=M2/M1

440/320=M2/44

1.45×44=M2

M2≈64