Answer:
The concentration of the solution is 5.8168 ×
mol.
Explanation:
Here, we want to calculate the concentration of the solution.
The unit of this is mol/dm^3
So the first thing to do here is to calculate the number of moles of the solute present, which is the number of moles of AlCO3
The number of moles = mass/molar mass
molar mass of AlCO3 = 27 + 12 + 3(16) = 27 + 12 + 48 = 87g/mol
Number of moles = 33.4/87 = 0.384 moles
This 0.384 moles is present in 660 L
x moles will be present in 1 dm^3
Recall 1 dm^3 = 1L
x * 660 = 0.384 * 1
x = 0.384/660 = 0.00058168 = 5.8168 * 10^-4 mol/dm^3
Answer:
Molar mass of solute: 300g/mol
Explanation:
<em>Vapor pressure of pure benzene: 0.930 atm</em>
<em>Assuming you dissolve 10.0 g of the non-volatile solute in 78.11g of benzene and vapour pressure of solution was found to be 0.900atm</em>
<em />
It is possible to answer this question based on Raoult's law that states vapor pressure of an ideal solution is equal to mole fraction of the solvent multiplied to pressure of pure solvent:

Moles in 78.11g of benzene are:
78.11g benzene × (1mol / 78.11g) = <em>1 mol benzene</em>
Now, mole fraction replacing in Raoult's law is:
0.900atm / 0.930atm = <em>0.9677 = moles solvent / total moles</em>.
As mole of solvent is 1:
0.9677× total moles = 1 mole benzene.
Total moles:
1.033 total moles. Moles of solute are:
1.033 moles - 1.000 moles = <em>0.0333 moles</em>.
As molar mass is the mass of a substance in 1 mole. Molar mass of the solute is:
10.0g / 0.033moles = <em>300g/mol</em>
Answer: 1800 L
Explanation:
Given that,
Original pressure of gas (P1) = 180 kPa
Original volume of gas (V1) = 1500 L
New pressure of gas (P2) = 150 kPa
New volume of gas (V2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P1V2
180 kPa x 1500 L = 150 kPa x V2
270000 kPa•L = 150 kPa•V2
Divide both sides by 150 kPa
270000 kPa•L/150 kPa = 150 kPa•V2/150 kPa
1800L = V2
Thus, the new volume of the gas is 1800 liters.