Step 1
The osmotic pressure is calculated as follows:

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Step 2
<em>Information provided:</em>
The mass of solute = 13.6 g
Volume of solution = 251 mL
Absolute temperature = T = 298 K
The molar mass of solute = M = 354.5 g/mol
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Step 3
Procedure:
1 L = 1000 mL => Volume = 251 mL x (1 L/1000 mL) = 0.251 L
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C = moles of solute/volume of solution (L)
C = mass of solute/(molar mass x Volume (L))
C = 13.6 g/(354.5 g/mol x 0.251 L)
C = 0.153 mol/L
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π = C x R x T
π = 0.153 mol/L x 0.082 atm L/mol K x 298 K
π = 3.74 atm
Answer: π = 3.74 atm
Answer:
(a) Moles of ammonium chloride = 0.243 moles
(b) 
(c) 60.68 mL
Explanation:
(a) Mass of ammonium chloride = 13.0 g
Molar mass of ammonium chloride = 53.491 g/mol
The formula for the calculation of moles is shown below:
Thus,
<u>Moles of ammonium chloride = 0.243 moles</u>
(b) Moles of ammonium chloride = 0.243 moles
Volume = 295 mL = 0.295 L ( 1 mL = 0.001 L)

(c) Moles of ammonium chloride = 0.0500 moles
Volume = ?
Molarity = 0.824 M
<u>Volume = 0.05 / 0.824 L = 0.06068 L = 60.68 mL</u>
I think it's easiest to find the pOH from the given [OH-] first.
-log(1x10^-5)
pOH=5
Then find the pH.
pOH+pH=14
5+pH=14
pH=9
Then find the [H+] using the pH.
antilog(-9) (if you dont have an antilog button use 10^-9)
[H+]=1x10^-9
Therefore is very important because it ensures survival, existence and stability of the environment.
2C6H14 + 19O2 ===> 12CO2 + 14H2O ... balanced equation
moled hexane present = 2.6 g x 1 mole/130 g = 0.02 moles
moles O2 present = 5.29 g x 1 mole/32 g = 0.165 moles
Which reactant is limiting? Hexane = 0.02/2 = 0.01; O2 = 0.165/19 = 0.0087
Thus O2 is limiting...
moles of H2O that can be produced =0.165 moles O2 x 14 H2O/19 CO2 = 0.122 moles H2O
Mass H2O = 0.122 moles x 18 gm/mole = 2.20 g (to 3 sig. figs.)