The osmotic pressure of a solution of 0.800 g of sucroseC12H22O11, in 150 g water at 35oC is:Question 9Select one:55 kPa46 kPa62 kPa48 kPa40 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 solute
• V is the volume of the solution 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,

n = 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 150 g of water is approximately 150 mL or 0.150 L.

Now we can substitute these values into the formula. We also need to convert the temperature to Kelvin by adding 273.15 to the Celsius temperature.

Π = 0.00234 mol / 0.150 L * 0.0821 L·atm/(K·mol) * (35°C + 273.15) K Π = 4.68 atm

To convert this pressure to kPa, we multiply by 101.325 (since 1 atm = 101.325 kPa).

Π = 4.68 atm * 101.325 kPa/atm = 474.4 kPa

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