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>
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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:
The answers are A,B,C.
Explanation: Just got it right on Edge 2020
Answer:
5.4 atm
Explanation:
P•V/T = P'•V'/T'
2.4•2.2/290 = P'•2/296
5.28/290 = P'/296 Multiply both sides by 296
1562.88/290 = P'
I'm = 5.4 atm
Answer: The volume of given gas is 2.27 L.
Explanation:
Given: Mass = 3.54 g
Temperature =
= (25 + 273) K = 298 K
Pressure = 725 mm Hg (1 mm Hg = 0.00131579) = 0.95 atm
As moles is the mass of substance divided by its molar mass.
So, moles of argon (molar mass = 40 g/mol) is as follows.

Formula used to calculate the volume of given gas is as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.

Thus, we can conclude that the volume of given gas is 2.27 L.
Answer:
239.15. OR round down 239
Explanation: