given:
Volume of
Cl2 gas V = 1 L
Temperature, T = 10°C
= 10 +273 (1°C = 273 K)
= 283 K
Pressure, P = 750 mm =
atm
Applying ideal gas equation, i.e
PV = nRT (i)
where n = Number of moles
R = Gas constant = 0.082 L atm
K−1mol From equation (i), we get
n =
Substituting all the values, we get
n =
= 0.0425
or,
Number of moles of
CI2 = 0.0425
Reaction between potassium iodide and
CI2 is:
2KI +
Cl2 → 2KCl +
I2 Now, from above reaction,
Moles of
CI2 = Moles of
I2 liberated
Thus, used moles of
I2 = 0.0425
Molecular mass of
I2 = 2 × 126.90
= 253.8
gmol−1 Now, number of moles is given by:
Number of moles =
| MolesofI2 |
| MolecularmassofI2 |
Substituting the values in above equation, we get
0.0425 =
Mass of
I2 liberated = 0.0425 × 253.8
= 10.7865 g
~ 10.78 g
Hence, option 'A' is correct.