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.)

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

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

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally