60.7 ml is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa.
Explanation:
Data given:
V1 = 75 ml
T1 = 30 Degrees or 273.15 + 30 = 303.15 K
P1 = 91 KPa
V2 =?
P2 = 1 atm or 101.3 KPa
T2 = 273.15 K
At STP the pressure is 1 atm and the temperature is 273.15 K
applying Gas Law:
=
putting the values in the equation of Gas Law:
V2 =
V2 =
V2 = 60.7 ml
at STP the volume of carbon dioxide gas is 60.7 ml.
Answer:
A) 2.69 M
B) 0.059
Explanation:
A) We have:
33.8% solute by mass= 33.8 g solute/100 g solution
molarity = mol solute/ 1 L solution
molarity= x
x
x
molarity= 2.69 mol solute/L solution = 2.69 M
B) We know that there are 33.8 g of solute in 100 g of solution.
As the total solution is compounded by solute+solvent (in this case, solvent is water), the mass of water is the difference between the mass of the total solution and the mass of solute:
mass of water= 100 g - 33.8 g = 66.2 g
Now, we calculate the number of mol of both solute and water:
mol solute= 33.8 g solute x = 0.232 mol
mol H20= 66.2 g H₂O x
Finally, the mol fraction of solute (Xsolute) is calculated as follows:
Xsolute=
Xsolute= 0.059