What mass of sucrose (C12H22O11) should be combined with 488 g of water to make a solution with an osmotic pressure of 8.00 atm at 290 K? (Assume the density of the solution to be equal to the density of the solvent.)

We need to add 51.52 grams of sucrose

Explanation:

Step 1: Data given

Mass of water = 488 grams

Density = 1g/mL

osmotic pressure = 8.00 atm

Temperature = 290 K

Step 2: Calculate concentration

π = i*M*R*T

8.00 atm = 1 * M*0.08206 * 290

M = 0.336 M

Step 3: Calculate volume of water

488 grams = 0.488 L

Step 4: Calculate moles sucrose

moles sucrose = molarity * volume

Moles sucrose = 0.336 M * 0.448 L

Moles sucrose = 0.1505 moles

Step 5: Calculate mass sucrose

Mass sucrose = 0.1505 moles * 342.3 g/mol

Mass sucrose = 51.52 grams

We need to add 51.52 grams of sucrose

56 g of sucrose is the mass needed

Explanation:

Formula for osmotic pressure → π = M. R. T

8 atm = M. 0.082 L. atm/mol. K. 290 K

8 atm / (0.082 L. atm/mol. K. 290 K) = M → 0.336 mol/L

Let's determine the mass of sucrose that represents 0.336 mol

0.336 mol. 342 g / 1mol = 114.9 g

This is the mass that corresponds to 1L of solution, but we have 0.488 L

Solution density = 1 g/mL → 488 g are contained in 488 mL.

488 mL. 1L / 1000 mL = 0.488 L

Let's make a rule of three:

1L is the volume for 114.9 g of sucrose

In 0.488 L of volume, we need a mass of (0.488L. 114.9 g) 1L = 56 g