Question

Define the following
(a) Leaching
(b) Metallurgy
(c) Anisotropy
Derive an expression for maximum work.
The boiling point of benzene is 353.23 K. When 1.80 gram of non-volatile solute was dissolved in 90 gram of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of solute.
[$K_b$ for benzene = 2.53 K mol${}^{-1}$]

Answer 1

Define the following
(a) Leaching
It is process of extracting a soluble material from an insoluble solid by dissolving out in a suitable solvent.
(b) Metallurgy
The process of extraction of a metal in a pure state from its ore is called metallurgy. Different methods include pyrometallurgy, hydrometallurgy and electrometallurgy.
(c) Anisotropy
The ability of crystalline solids to change values of physical properties when measured in different directions is called anisotropy.
An expression for maximum work:
When the volume of gas increases by infinitesimal amount dV in a single step, the small quantity of work done is given by $dW=-PdV$
Total amount of work done during expansion from volume V to V' is maximum work and is obtained by integration.

$W_{max}{\int }_1^{2}dW=-{\int }_V^{V^{\prime }}PdV$
Substitute $P=\frac{nRT}{V}$
$W_{max}{\int }_1^{2}dW=-{\int }_V^{V^{\prime }}\frac{nRT}{V}dV$
For isothermal process T is constant
$W_{max}=-nRT{\int }_V^{V^{\prime }}\frac{dV}{V}$
$W_{max}=-nRTln\frac{V^{\prime }}{V}$
Calculation for the molar mass of solute:
$M_2=\frac{K_b{W}_2}{\Delta T_b{W}_1}$
Here, $M_2$ is the molar mass of solute, $K_b$ is the molal elevation in boiling point constant, $W_2$ is the mass of solute, $\Delta T_b$ is the elevation in the boiling point, $W_1$ is the mass of the solvent.
$M_2=\frac{2.53\times 1.80}{(354.11-353.23)\times 90}$
$M_2=0.0575$ kg/mol
$M_2=57.5$ g/mol
Hence, the molar mass of the solute is 57.5 g/mol.