Given that CaCl2 = 2.46 m
Therefore 2.46 mole of CaCl2 present in 1.00 g of solution
Mass of 2.46 mole CaCl2= number of mole * molar mass
= 2.46 mole * 110.98 g/mol
= 273.01 g
Mass of water or solvent = mass of solution – mass of solute
= 1000 g -273.01 g
= 726.99 g
Mole of water = amount in g / molar mass
= 726.99 g/18.02 g/ mole
= 40.34 moles
Mole fraction of CaCl2 = number of mole of CaCl2 / total moles
= 2.46 /2.46+40.34
= 2.46/42.8
= 0.057
Answer : The work done by the system is, 2.2722 J
Explanation :
The expression used for work done in reversible isothermal expansion will be,
where,
w = work done = ?
n = number of moles of gas = 0.00100 mole
R = gas constant = 8.314 J/mole K
T = temperature of gas =
= initial volume of gas = 25 mL
= final volume of gas = 75 mL
Now put all the given values in the above formula, we get:
Therefore, the work done by the system is, 2.2722 J
Answer:
34.28 L ( 1.5*22.4 L)
Explanation:
Calculation of the moles of aluminum as:-
Mass = 55 g
Molar mass of aluminum = 26.981539 g/mol
The formula for the calculation of moles is shown below:
Thus,
According to the reaction:-
4 moles of aluminum react with 3 moles of oxygen gas
1 mole of aluminum react with moles of oxygen gas
2.0384 moles of aluminum react with moles of oxygen gas
Moles of oxygen gas = 1.5288 moles
At STP,
Pressure = 1 atm
Temperature = 273.15 K
Using ideal gas equation as:
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 = 0.0821 L.atm/K.mol
Applying the equation as:
1 atm × V = 1.5288 mol × 0.0821 L.atm/K.mol × 273.15 K
⇒V = 34.28 L ( 1.5*22.4 L)