What mass of sucrose should be combined with 524 g of water to make a solution with an osmotic pressure of 8.50 atm at 270 K? (Assume the density of the solution to be equal to the density of the solvent.)

To find the mass of sucrose needed to make the solution, we can use the formula for osmotic pressure:

π = (n/V)RT

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/(mol·K)),
and T is the temperature in Kelvin.

First, let's calculate the volume of the solution in liters. Since the density of the solution is assumed to be equal to the density of the solvent (water), we can use the relationship between mass, volume, and density:

density = mass/volume

Rearranging the equation, we have:

volume = mass/density

The density of water is approximately 1 g/mL or 1 g/cm³. Since the mass provided is in grams, we can use the same units for the density. So, the volume of the solution is:

volume = 524 g / (1 g/cm³) = 524 cm³ (or 524 mL)

Since 1 cm³ is equal to 1 mL, we can use either unit in the subsequent calculations.

Next, we need to convert the temperature from Celsius to Kelvin. The conversion formula is:

T(K) = T(°C) + 273.15

T = 270 °C + 273.15 = 543.15 K

Now, we can rearrange the osmotic pressure formula to solve for the number of moles of sucrose:

n = (πV) / (RT)

Substituting the provided values, we get:

n = (8.50 atm * 524 mL) / (0.0821 L·atm/(mol·K) * 543.15 K)

n ≈ 66.74 mol

The molar mass of sucrose (C12H22O11) is approximately 342.3 g/mol. To find the mass of sucrose needed, we can use the equation:

mass = moles * molar mass

mass = 66.74 mol * 342.3 g/mol

mass ≈ 22,861 g

Therefore, approximately 22,861 grams of sucrose should be combined with 524 grams of water to make a solution with an osmotic pressure of 8.50 atm at 270 K.