Question

What is osmotic pressure of a solution of bovine insulin (molar mass, 5700 g mol^-1) at 18 °C if 100.0 mL of the solution contains 0.103 g of the insulin?

Answer 1

3.28mmHg

Explanations:

According to ideal gas equation

`PV=nRT`

P is the pressure (in atm)

• V is the ,volume (i,n L)

,

• n is the ,moles, of the insulin

,

• R is the ,Gas constant = 0.08205Latm/molK

,

• T is the ,temperature

Determine the moles of insulin

moles = mass/molar mass

moles of insulin = 0.103/5700

moles of insulin = 0.00001807moles

Determine the required pressure

`\begin{gathered} P=(nRT)/(V) \\ P=(0.00001807*0.08205*(18+273))/(0.1L) \\ P=(0.0004315)/(0.1) \\ P=0.004315atm \end{gathered}`

Convert the unit to mmHg

Using the conversion factor

1atm = 760mmHg

0.004315atm = 0.004315atm * 760

0.004315atm = 3.28mHg

Therefore the osmotic pressure of a solution is 3.28mmHg