When water was added to a 4.00 gram mixture of potassium oxalate hydrate (molar mass 184.24 g/mol) and calcium hydrate shown below , 1.20 g of calcium oxalate hydrate (146.12 g/mol molar mass) was recovered. if the mole: mole ratio of potassium oxalate to calcium oxalate is 1: 1 , what percentage of the 4g-mixture is potassium oxalate

% (COOK)2H2O = 37.826 %

Explanation:

mix: (COOK)2H2O + Ca(OH)2 → CaC2O4 + H2O

∴ mass mix = 4.00 g

∴ mass (CaC2O4)H2O = 1.20 g

∴ Mw (COOK)2H2O = 184.24 g/mol

∴ Mw (CaC2O4)H2O = 146.12 g/mol

∴ r = mol (COOK)2H2O / mol (CaC2O4)H2O = 1

⇒ mass (COOK)2H2O = (1.20 g (CaC2O4)H2O)×(mol (CaC2O4)H2O / 146.12 g (CaC2O4)H2O)×(mol (COOK)2H2O/mol (CaC2O4)H2O)×(184.24 g (COOK)2H2O/mol (COOK)2H2O)

⇒ mass (COOK)2H2O = 1.513 g

⇒ % (COOK)2H2O = ( 1.513 g / 4 g )×100

⇒ % (COOK)2H2O = 37.826 %

π = 4,882 atm

Explanation:

To calculate the osmotic pressure (π), the Van´t Hoff equation must be used, which is:

π x V = n x R x T

Where:

• π: Osmotic pressure, which is the difference between the levels of the solution and the pure solvent through a semipermeable membrane, which allows the passage of the solvent but not the solute

• V: Volume of the solution, in liters unit

• n: Number of moles of solute

• R: Constant of ideal gases, equal to 0.08206 L.atm / mol.K

• T: Absolute temperature, in Kelvin degrees

With the data you provide you can calculate the osmotic pressure by clearing it from the equation, we would be equal to:

π = (n x R x T) / V

However, all data must first be converted to the corresponding units in order to replace the values in the equation.

Solution volume ⇒ go from mL to L:

1000 mL of solution 1 L

2600 mL of solution X = 2.6 L

Calculation: 2600 mL x 1 L / 1000 mL = 2.6 L

Temperature ⇒ Go from ° C to K

T (K) = t (° C) + 273.15 = 30.0 ° C + 273.15 = 303.15 K

Number of moles of solute ⇒ It can be calculated since we have the mass of the enzyme and its molecular mass:

98.0 g of enzyme 1 mol

50.0 g of enzyme X = 0.510 moles

Calculation: 50.0 g x 1 mol / 98.0 g = 0.510 moles

Now, you can replace the values in the Van´t Hoff equation and you will get the result:

 π = (n x R x T) / V

π = (0.510 mol x 0.08206 L.atm / mol.K x 303.15 K) / 2.6 L = 4.882 atm

Therefore, the osmotic pressure will be 4,882 atm