Answer:
MM = 5,521.54 g/mol
Explanation:
To solve this, we need to use the expression for osmotic pressure which is the following:
π = MRT (1)
Where:
M: Concentration of the solution
R: gas constant (0.082 L atm/ mol K
T: temperature in K
25 °C in Kelvin is: 25 + 273.15 = 298.15 K
Now, we do not have the concentration of the solution, but we do have the mass. and the concentration can be expressed in terms of mass, molar mass and volume:
Concentration (M) is:
M = n/V (2)
and n (moles) is:
n = m/MM (3)
Therefore, if we replace (2) and (3) in (1) we have:
π = mRT/V*MM
Solving for MM we have:
MM = mRT/πV (4)
All we have to do now, is replace the given data and we should get the value of the molar mass:
MM = 6.143 * 0.082 * 298.15 / 0.1 * 0.272
MM = 150.1859 / 0.0272
<em>MM = 5,521.54 g/mol</em>
<em>This is the molar mass.</em>
Answer:
The answer to your question is V2 = 434.7 l
Explanation:
Data
Volume 1 = V1 = 240 l Volume 2 = ?
Temperature 1 = T1 = 479°K Temperature 2 = T2 = 293°K
Pressure 1 = P1 = 300 KPa Pressure 2 = P2 = 101.325 Kpa
Process
1.- Use the combined gas law to solve this problem
P1V1/T1 = P2V2/t2
-Solve for V2
V2 = P1V1T2 / T1P2
2.- Substitution
V2 = (300)(240)(293) / (479)(101.325)
3.- Simplification
V2 = 21096000 / 48534.675
4.- Result
V2 = 434.7 l
Answer:
8 moles of C
Explanation:
From the question given above, the following equation was obtained:
3A + 2B —> 6C
From the equation above,
3 moles of A reacted to produce 6 moles of C.
Thus, the number of mole of C produced by reacting 4 moles of A can be obtained as follow:
From the equation above,
3 moles of A reacted to produce 6 moles of C.
Therefore, 4 moles of C will react to produce = (4 × 6)/3 = 8 moles of C
Thus, 8 moles of C can be obtained from the reaction of 4 moles of A with excess B
