Question

A current of dry air was passed first through a series of bulbs containing a solution of C6H5NO2 in ethanol of molality 0.725 and then through a series of bulbs containing pure ethanol (T = 284k). Loss in weight of the solvent bulbs was 0.0685g. Calculate the loss of weight in the solution bulbs.

Answer 1

A current of dry air was passed first through a series of bulbs containing a solution of C6H5NO2 in ethanol of molality 0.725 and then through a series of bulbs containing pure ethanol (T = 284k). Loss in weight of the solvent bulbs was 0.0685g. Calculate the loss of weight in the solution bulbs.

ANSWER0. 5555

Answer 2

The loss of weight in the solution bulbs will be Ps = $2.05 gm$

Explanation:

• no of moles of solute $=n1$
• no of moles of solvent $=n1$
• $\frac{n2}{n1}$=$\frac{mM}{1000}$
• where m is the molality of the solution.
• M is the molar mass of ethanol.
• given the solution of ethanol, $C2H5OH$  Its molar mass is equal to 46
• $\frac{n2}{n1}= \frac{(0.725)(46)}{1000}$
• [tex]=0.033 [/tex]
• According to Raoult's law
• $\frac{Po -Ps}{Ps}=\frac{n2}{n1}$ -------- 1
• $Po=$ vapor pressure of the solution.
• $Ps =$ vapor pressure of the solvent.
• we know that $Ps$ is proportional to the loss in weight of the solution.
• Loss in weight of the solvent bulbs= 0.0685g
• Substituting the values in Eqn 1 we get
• $\frac{0.0685}{Ps}= 0.0334$
• Ps = $2.05 gm$