Answer:
85.7°C
Step-by-step explanation:
Step 1:
Data obtained from the question. This include the following:
Volume (V) = 73.3mL = 73.3/1000 = 0.0733L
Mass of l2 = 0.292g
Pressure (P) = 0.462atm
Temperature (T) =..?
Step 2:
Determination of the number of mole present in 0.292g of I2. This is illustrated below:
Mass of l2 = 0.292g
Molar Mass of I2 = 127 x 2 = 254g/mol
Number of mole of I2 =..?
Number of mole = Mass /Molar Mass
Number of mole of I2 = 0.292/254
Number of mole of I2 = 1.15×10¯³ mole
Step 3:
Determination of the temperature.
The temperature in the bulb containing the iodine vapor can be obtained by using the ideal gas equation as follow:
Volume (V) = 0.0733L
Pressure (P) = 0.462atm
Number of mole (n) = 1.15×10¯³ mole
Gas constant (R) = 0.0821 atm.L/Kmol
Temperature (T) =..?
PV = nRT
0.462 x 0.0733 = 1.15×10¯³ x 0.0821 x T
Divide both side by 1.15×10¯³ x 0.0821
T = (0.462 x 0.0733)/(1.15×10¯³ x 0.0821)
T = 358.7K
Converting 358.7K to celsius temperature, we have:
T(°C) = T(K) – 273
T(K) = 358.7K
T(°C) = 358.7 – 273
T(°C) = 85.7°C
Therefore, the temperature is 85.7°C