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

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

indium, and thallium have three electrons i think