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121.20OmL of an aqueous solution of a protein contains its 1.26 g The osmotic pressure of this solution at300 K is found to be 2.57 x 10-3 bar. The molar mass of prote in will be (R=0.083Lbarmol-lk-l) (a)S1022gmol (b)122044gmol- (c)3 101lg mol (d)61038g mol​

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Answer:

Given data:

Volume of solution, V = 121.20 mL = 0.1212 L

Mass of protein, m = 1.26 g

Osmotic pressure, π = 2.57 x 10^-3 bar

Temperature, T = 300 K

Universal gas constant, R = 0.083 L bar mol^-1 K^-1

We can use the formula for osmotic pressure: π = (n/V)RT, where n is the number of moles of solute.

First, we need to calculate the number of moles of protein: n = m / M, where M is the molar mass of the protein.

Next, we can rearrange the formula for osmotic pressure to solve for the molar mass, M: M=m/n

Substitute the values into the equations: n = 1.26 g / M π = (n / V)RT

Now, rearrange the osmotic pressure formula to solve for M: M = m / (πV/RT)

Substitute the given values into the equation and solve for M: M=1.26 g/((2.57 x 10^-3 bar)* 0.1212 L/ (0.083 L bar mol^-1 K^-1* 300 K))

Calculating the value of M will give the molar mass of the protein in grams per mole.

Therefore, the molar mass of the protein will be calculated to

determine which option it matches with:

(a) 51022 g/mol

(b) 122044 g/mol

(c) 31011 g/mol

(d) 61038 g/mol

Please calculate the final answer accordingly.

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