Solution:
7) According to Graham's law of effusion,
t1/t2 = √(M1/M2)
Where,
t1 = 43 sec
M1 = molecular mass of N2O = 44 g /mol
M2 = molecular mass of I2 = 254 g /mol
t2 = ?
Hence,
43 sec / t2 = √( 44 / 254)
t2 = √(254/44) x 43 sec = √5.77 x 43 sec
t2 = 2.40 x 43 sec = 103.2 sec
t2 = 100 sec ( in two significant figure)
8) According to ideal gas equation,
PV = nRT
V = n (RT/P)
Initial:
249 mL = 0.518 mol x (RT/P) ----(1)
Final:
V = (0.518 + 0.311) mol x (RT/P) ----(2)
Dividing equation 2 by 1,
V/249 mL = 0.829 / 0.518
V = 249 mL x 0.829/0.518 = 398.5 mL
Q78 Part A How long would it take the same amount of gaseous I, to effuse...
A sample of N2O effuses from a container in 49 seconds. How long would it take the same amount of gaseous I2 to effuse from the same container under identical conditions?
A sample of N2O effuses from a container in 42 seconds. How long would it take the same amount of gaseous I2 to effuse from the same container under identical conditions?
A sample of N2O effuses from a container in 49 seconds. How long would it take the same amount of gaseous I2 to effuse from the same container under identical conditions?
A sample of N2O effuses from a container in 46 seconds. How long would it take the same amount of gaseous I2 to effuse from the same container under identical conditions? Express your answer using two significant figures.
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