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

Answer:

The value of n is 14.

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 =$\frac{57.1}{760} atm = 0.07513 atm$

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:

$c=\frac{\pi}{iRT}=\frac{0.07513 atm}{2\times 0.0821 atm L/mol K\times 298.15 K}$

$c=0.001535 mol/L$

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

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

Molality of the solution = m = 0.001535 mol/L

$\frac{0.036 g}{M\times 0.1 kg}=0.001535 mol/kg$

M = 234.53 g/mol

Molar mass of $LiC_nH_{2n+1}O_2$ : M

M = $7 g/mol\times 1 + 12 g/mol \times n +1 g/mol\times (2n+1)+2\times 16 g/mol$

$234.53 g/mol=7 g/mol\times 1 + 12 g/mol \times n +1 g/mol\times (2n+1)+2\times 16 g/mol$

n = 14

The value of n is 14.