Question

A starch has a molar mass of 3.20 x 104 g/mol. If 0.759 g of this starch is dissolved in 112 mL of solution, what is the osmotic pressure, in torr, at 25.00 oC?

Answer 1

The osmotic pressure is 3.94 torr.

Step-by-step explanation:

Osmotic pressure is a colligative property of solution, which can be calculated according to the following expression:

π = M . R. T [1]

where,

π is the osmotic pressure

M is the molarity of the solution

R is the ideal gas constant (62.36 L . torr/ K . mol)

T is the absolute temperature (in Kelvin scale)

Molarity can be calculated from its definition:

`M=\frac{moles \hspace{2} of solute}{litres\hspace{2} of solution} =\frac{mass\hspace{2} of solute}{molar \hspace{2} mass\hspace{2} of solute\hspace{2} . litres \hspace{2}of \hspace{2}solution} = (0.759g)/(3.20.10^(4)g/mol . 0,112L ) =2.12.10^(-4) M`

Absolute temperature can be calculated like:

K = °C + 273.15

K = 25.00 + 273.15 = 298.15 K

Replacing this data in equation 1:

`\pi=2.12.10^(-4) (mol)/(L) .62.36(L.torr)/(K.mol).298.15K= 3.94torr`