Question

Q78

Part A How long would it take the same amount of gaseous I, to effuse from the same container under identical conditions? Express your answer using two significant figures. IVO AED 6 0 ? Submit Request Answer

Answer 1

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