What is the pOH of a 5.6 x 10-5 M solution of cesium hydroxide (CsOH)?
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Answer:
Explanation:
Given that:
Pressure = 791 mmHg
Temperature = 20.0°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (20 + 273.15) K = 293.15 K
T = 293.15 K
Volume = 100 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 62.3637 L.mmHg/K.mol
Applying the equation as:
791 mmHg × 1.14 L = n × 62.3637 L.mmHg/K.mol × 293.15 K
⇒n of produced = 0.0493 moles
According to the reaction:-
1 mole of carbon dioxide is produced 1 mole of calcium carbonate reacts
0.0493 mole of carbon dioxide is produced 0.0493 mole of calcium carbonate reacts
Moles of calcium carbonate reacted = 0.0493 moles
Molar mass of = 100.0869 g/mol
The formula for the calculation of moles is shown below:
Thus,
Impure sample mass = 5.28 g
Percent mass is percentage by the mass of the compound present in the sample.