The osmotic pressure of a solution of 0.800 g of sucrose C12H22O11,in 100g water at 27oC is:Question 9Select one:63 kPa52 kPa58 kPa42 kPa46 kPa

To solve this problem, we need to use the formula for osmotic pressure:

Π = n/V * R * T

where:

• Π is the osmotic pressure,
• n is the number of moles of the solute,
• V is the volume of the solvent in liters,
• R is the ideal gas constant (0.0821 L·atm/(K·mol) or 8.3145 J/(K·mol) depending on the units of pressure),
• T is the temperature in Kelvin.

First, we need to convert the mass of sucrose to moles. The molar mass of sucrose (C12H22O11) is approximately 342.3 g/mol. So, 0.800 g of sucrose is 0.800 g / 342.3 g/mol = 0.00234 mol.

Next, we need to convert the mass of water to volume. The density of water is approximately 1 g/mL, so 100 g of water is 100 mL or 0.1 L.

The temperature needs to be converted to Kelvin. T = 27°C + 273.15 = 300.15 K.

Now we can substitute these values into the formula. If we use R = 0.0821 L·atm/(K·mol), the units of pressure will be atmospheres. If we use R = 8.3145 J/(K·mol), the units of pressure will be Pascals. Since the answer choices are in kPa, we'll use the second value for R.

Π = (0.00234 mol / 0.1 L) * 8.3145 J/(K·mol) * 300.15 K = 582.6 Pa = 0.5826 kPa

None of the answer choices match this result. There may be a mistake in the problem or the answer choices.