What is the osmotic pressure of a solution made by dissolving 55.0 g of glucose, C6H12O6, in enough water to form 800.0 mL of solution at 10.0 ∘C ? Express your answer to three significant figures and include the appropriate units. View Available Hint(s)

The osmotic pressure is 8.85 atm

Explanation:

Step : Data given

Mass of glucose = 55.0 grams

Volume of water =800.0 mL

Temperature = 10.0 °C

Step 2: Calculate moles glucose

Moles glucose = mass glucose / molar mass glucose

Moles glucose = 55.0 grams / 180.156 g/mol

Moles glucose = 0.305 moles

Step 3: Calculate osmotic pressure

π = iMRT  

⇒ with π = the osmotic pressure = TO BE DETERMINED

⇒ with i = the van't Hoff factor for glucose = 1

⇒ with M = the concentration = moles / volume = 0.305 moles / 0.800 L =0.381 M

⇒ with R = the gas constant = 0.08206 L*atm*mol¨K

⇒ with T = the temperature = 10.0 °C = 283 K

 π = 1* 0.381 *0.08206 * 283

 π = 8.85 atm

The osmotic pressure is 8.85 atm

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.