A sample 21.5 mg of the protein was dissolved in water at 15.0 degree C to make 2.46 mL of...

Question:

A sample {eq}21.5 \ mg {/eq} of the protein was dissolved in water at {eq}15.0^\circ C {/eq} to make {eq}2.46 \ mL {/eq} of solution and measures an osmotic pressure of {eq}3.92 \ torr {/eq} .What is the molar mass of this protein?

Known value {eq}R = 0.08206 \ atm \cdot L / mol \cdot K. {/eq}

Molar mass:

The molar mass of a compound is defined as the sum of masses of all the elements that form a compound. It is a very important parameter in order to calculate all other stoichiometric properties and values of it.

STEP 1:

The given data are:

{eq}w= 21.5 \enspace mg {/eq}

{eq}T=15^{\circ}c = 15+273= 288 \enspace k {/eq}

{eq}V= 2.46 \enspace mL = 0.00246 \enspace L {/eq}

{eq}P= 3.92 \enspace torr = \dfrac{3.96}{760} = 0.0052 \enspace atm {/eq}

STEP 2:

The molar mass can be calculated as:

We can write ideal gas equation as:

{eq}PV=nRT {/eq}

Where {eq}n =\dfrac{w}{M} {/eq}

Substituting in above equation we get:

{eq}PV= \dfrac{wRT}{M} {/eq}

{eq}M= \dfrac{wRT}{PV} {/eq}

{eq}M= \dfrac{0.0215 \times 0.0821 \times 288}{0.0052 \times 0.00246} = 39740.7 \enspace g/mol {/eq}

Therefore, the molar mass of the protein comes out to be {eq}39740.7 \enspace g/mol {/eq} 