79. At 300 K and 1 atm, 15mL of a gaseous hydrocarbon requires 375 mL air
containing 20% O2 by volume for complete combustion. After combustion the
gases occupy 330 mL. Assuming that the water formed is in liquid form and the
volumes were measured at the same temperature and pressure, the formula of the
hydrocarbon is:
(A)
C4 H8
(B)
C4 H10
(C)
C3 H6
(D)
C3 H8
Solution: (B)
Volume of N2 in air = 375 × 0.8 = 300 ml
Volume of O2 in air = 375 × 0.2 = 75 ml
After combustion total volume
330 = VN2 + VCO2
330 = 300 + 15x
x=2
Volume of O2 used
y
15 (x + ) = 75
4
y
x+ =5
4
y = 12
So hydrocarbon is = C2 H12
None of the option matches it therefore it is a BONUS.
Alternatively
Volume of O2 used
y
15 (x + ) = 75
4
y
x+ =5
4
If further information (i.e., 330 ml) is neglected, option (C3 H8 ) only satisfy the
above equation.