Question

A 0.1 (M) solution of an acid HA of density 1.01 gm/ml is taken in a beaker. The molecular mass of HA is 300 $g/mol$. The freezing point of solution is $0.2087{}^{o}C$. For $H_{2}O$, $K_f=1.86K/m$

What is the degree of dissociation of HA in its 0.1(M) aqueous solution (in percentage) ?

• A. $10$
• B. $20$
• C. $25$
• D. $55.56$

Answer 1

The correct option is A $10$

we have,

$\Delta T_f=i{K}_{f}m$

density of solution = 1.01 gm/ml

volume of solution = 1000 ml

mass of solution = density * volume = 1010 g

moles of solute = 0.1 moles

molar mass = 300

so, mass of solute = 0.1 * 300 = 30 g

mass of solvent = mass of solution - mass of solute = 1010 - 30 = 980g

so , putting the values in

$\Delta T_f=i{K}_{f}m$

we have,

$0.2087=i\ast 1.86\ast \frac{0.1\ast 1000}{980}$

i = 1.09

and we have

$i=1+\alpha (n-1)$

$1.09=1+\alpha (2-1)$

$\alpha =.09or10$