# An aqueous solution containing 1.00 g of bovine insulin (a protein, not ionized) per liter has an osmotic pressure of 3.1 mm Hg at 25 °C. How do you calculate the molar mass of bovine insulin?

May 31, 2018

The molar mass is 6.0 ×10^3color(white)(l)"g/mol".

#### Explanation:

The formula for osmotic pressure $\Pi$ is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \Pi = c R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where

$c =$ the molar concentration of the solute
$R =$ the universal gas constant
$T =$ the Kelvin temperature of the solution

Since $c = \text{moles"/"litres} = \frac{n}{V}$, we can write

$\Pi = \frac{n R T}{V}$

Since $n = \text{mass"/"molar mass} = \frac{m}{M}$, we can write

$\Pi = \frac{n R T}{M V}$

We can solve this equation for the molar mass and get

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} M = \frac{m R T}{\Pi V} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

In this problem

$m = \text{1.00 g}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{25 °C = 298.15 K}$
Pi = 3.1 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "0.004 08 atm"
$V = \text{1 L}$

M = ("1.00 g" × "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 298.15 color(red)(cancel(color(black)("K"))))/("0.00 408"color(red)(cancel(color(black)("atm"))) × 1 color(red)(cancel(color(black)("L")))) = 6.0 × 10^3color(white)(l)"g/mol"