What is the osmotic pressure of an intravenous solution prepared by dissolving 44.0 g of dextrose (sugar), C6H12O6, in enough water to make 1.00 L of solution? Dextrose is a nonelectrolyte and body temperature is 37.0 ∘C.

6.20atm

Explanation:

Volume = 1L

Mass of dextrose (C6H12O6) = 44g

First, let us calculate the molarity of dextrose sugar. This is illustrated below:

Molar Mass of dextrose (C6H12O6) = (12x6) + (12x1) + (16x6) = 72 + 12 + 96 = 180g/mol

Number of mole of dextrose = Mass / Molar Mass = 44/180 = 0.244mol

Molarity of dextrose = mole /Volume

Molarity of dextrose = 0.244/1

Molarity = 0.244mol/L

Using the osmotic pressure formula, we can simply solve as illustrated below:

Π = iMRT

i = 1 ( since dextrose is non electrolyte)

M = 0.244mol/L

R = 0.082atm.L/K /mol

T = 37°C = 37 + 273 = 310K

Π =?

Π = iMRT

Π = 1 x 0.244 x 0.082 x 310

Π = 6.20atm

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

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