Question

A lithium salt used in lubricating grease has the formula LiCnH2n+1O2. The salt is soluble in water to the extent of 0.036 g per 100 g of water at 25 ∘C. The osmotic pressure of this solution is found to be 57.1 torr. Assuming that molality and molarity in such a dilute solution are the same and that the lithium salt is completely dissociated in the solution, determine an appropriate value of n in the formula for the salt.

Answer 1

The value of n is 14.

Step-by-step explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

`\pi=icRT`

where,

`\pi` = osmotic pressure of the solution = 57.1 Torr =
`(57.1)/(760) atm = 0.07513 atm`

1 atm = 760 Torr

i = Van't hoff factor = 2 (electrolytes)

c = concentration of solute = ?

R = Gas constant =
`0.0820\text{ L atm }mol^(-1)K^(-1)`

T = temperature of the solution =
`25^oC=[273.15+25]=298.15 K`

Putting values in above equation, we get:

`c=(\pi)/(iRT)=(0.07513 atm)/(2* 0.0821 atm L/mol K* 298.15 K)`

`c=0.001535 mol/L`

Assuming that molality and molarity in such a dilute solution.

c = m (Molality)

The salt is soluble in water to the extent of 0.036 g per 100 g of water at 25°C

`Molaity=\frac{\text{Mass of solute}}{\text{molar mass of solute(M)}* \text{Mass of solvent in kg}}`

Molality of the solution = m = 0.001535 mol/L

`(0.036 g)/(M* 0.1 kg)=0.001535 mol/kg`

M = 234.53 g/mol

Molar mass of
`LiC_nH_(2n+1)O_2` : M

M =
`7 g/mol* 1 + 12 g/mol * n +1 g/mol* (2n+1)+2* 16 g/mol`

`234.53 g/mol=7 g/mol* 1 + 12 g/mol * n +1 g/mol* (2n+1)+2* 16 g/mol`

n = 14

The value of n is 14.