If 1.30 grams of a protein (non -electrolyte) are added to a 100.0 mL of water, the osmotic pressure of the resulting solution is 0.0074 atm at 298 K. What is the molar mass of the protein ?

M = 0.43× 10⁵ g/mol

Explanation:

Given data:

Mass of protein = 1.30 g

Volume of water = 100.0 mL ( 100/1000 =0.1 L)

Pressure of solution = 0.0074 atm

Temperature = 298 K

Molar mass of protein = ?

Solution:

Formula:

PV = nRT

n = PV / RT

n = 0.0074 atm ×0.1 L / 0.0821 atm. L / mol.K × 298 K

n = 0.00074 / 24.5 /mol

n = 3.0 × 10⁻⁵ mol

Molar mass of protein:

Number of moles = mass /  molar mass

3.0 × 10⁻⁵ mol = 1.30 g/ M

M = 1.30 g/  3.0 × 10⁻⁵ mol

M = 0.43× 10⁵ g/mol

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2