# What is the osmotic pressure for #"1.95 g"# of sucrose dissolved in #"150 mL"# solution at #25^@ "C"#? #"FW"# #=# #"342.2965 g/mol"#

##### 1 Answer

#Pi = "0.9292 atm"# .

You can read more about osmotic pressure (and other colligative properties) here.

The **osmotic pressure**

#Pi = icRT# ,where:

#i# is thevan't Hoff factor, as seen in freezing point depression and boiling point elevation. It can be approximated by the number of particles per formula unit in solution.#c# is theconcentrationin whatever units are appropriate. In this case, if#Pi# is in#"atm"# , then#c# is in#"mol/L"# when...- ...the
universal gas constant#R# is#"0.082057 L"cdot"atm/mol"cdot"K"# .#T# is thetemperaturein#"K"# .

We are indeed given the correct info to calculate the molarity of the sucrose solution.

#1.95 cancel"g suc" xx "1 mol suc"/(342.2965 cancel"g suc")#

#=# #"0.005697 mols suc"#

So, the molarity is:

#c = "0.005697 mols suc"/(150 xx 10^(-3) "L soln")#

#=# #"0.03798 M"#

Therefore, the osmotic pressure at

#color(blue)(Pi) = (1)(0.03798 cancel"mol/L")(0.082057 cancel"L"cdot"atm/"cancel"mol"cdotcancel"K")(298.15 cancel"K")#

#=# #color(blue)("0.9292 atm")# ,

since nonelectrolytes have