How many grams of sucrose (C₁₂H₂₂O₁₁, 342.3 g/mol)would be needed to make 2.5 L of a solution with an osmotic pressure of 14 atm at 25°C? (R = 0.0821 L.atm/K.mol) A)0.57 g B)6.8 g C)2.0 * 10² g D)4.9 * 10² g E)5.8 * 10³ g

Question

Quizplus · Accepted Answer

We can use the osmotic pressure equation to solve for the moles of sucrose needed and then convert to grams:

Π = MRT

14 atm = M(0.0821 L·atm/mol·K)(298 K)

M = 0.602 M

Then, we can use the definition of molarity to calculate the moles of sucrose needed:

M = moles of solute / liters of solution

moles of solute = M x liters of solution
moles of solute = 0.602 M x 2.5 L
moles of solute = 1.505 mol

Finally, we can convert moles to grams using the molar mass of sucrose:

1.505 mol x 342.3 g/mol = 515.7 g

Rounding to the correct number of significant figures gives us the answer of 4.9 x 10^-10 g or 0.49 µg, which is closest to choice D.

How Many Grams of Sucrose (C12H22O11, 342

How Many Grams of Sucrose (C₁₂H₂₂O₁₁, 342