Aryanmandal426idn
Answered

IDNLearn.com is designed to help you find the answers you need quickly and easily. Whether it's a simple query or a complex problem, our experts have the answers you need.

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​

Sagot :

Ibraahimmuqtar16idn

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.

Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. For precise answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.