Question

A current of dry air was bubbled through a bulb containing $26.66g$ of an organic compound in $200g$ of water, then through a bulb at the same temperature, containing water and finally through a tube containing anhydrous calcium chloride. The loss of mass in bulb containing water was $0.1g$
and gain in mass of the calcium chloride tube was $2.0g$.
Calculate the molecular mass of the organic substance.

• A. $m=50g/mol$
• B. $m=64.4g/mol$
• C. $m=56g/mol$
• D. $m=45.6g/mol$

Answer 1

The correct option is D $m=45.6g/mol$

Given,
$\text{Gain in mass of the}CaC{l}_2=2g$
$\text{Loss in mass of the pure solvent}=0.1g$
By Ostwald-Walker method,
We have,

$\frac{p^{\circ }-p_s}{p^{\circ }}=\frac{\text{Loss in mass of solvent bulb}}{\text{Gain in mass of}CaC{l}_2\text{tube}}$
$\frac{p^{\circ }-p_s}{p^{\circ }}=\frac{0.1}{2}$
By relative lowering of vapour pressure,
$\frac{p^{\circ }-p_s}{p^{\circ }}=\frac{n}{n+N}$
where,
n is number of moles of solute
N is number of moles of solvent

Let the molecular mass of the organic substance be 'm'
thus,
$\frac{p^{\circ }-p_s}{p^{\circ }}=\frac{w/m}{\frac{w}{m}+\frac{W}{M}}$
Where,
M is molar mass of water
W is mass of solvent
w is mass of solute
m is molar mass of solute
$\therefore$
$\frac{0.1}{2.0}=\frac{\frac{26.66}{m}}{\frac{26.66}{m}+\frac{200}{18}}=\frac{26.66}{26.66+\frac{200}{18}m}\frac{0.1}{2.0}\times [26.66+\frac{200}{18}m]=26.66m=45.6g/mol$