Question

The vapour pressure of an aqueous solution of sucrose at 373k is found to be 750mm hg. the molality at same temperature is

Answer 1

Given that  

pressure at 373 of solution is 750 mm Hg.  

Pressure of solvent means water at 373 K is 760 mmHg

According to the Raoult’s law;

Psoln = PH2O XH2O  

750 = 760 (XH2O)  

mole fraction; XH2O = 0.987  

mole fraction = moles of water / total number of moles (water + solute )

Here mass of water is 1.00 kg or 1000 g

Therefore;  

Number of moles = amount in g / molar mass

= 1000g  18.02 g/ mole=  55.6 moles H2O,  

And from mole frcation;

0.987 = 55.6 /(55.6 + x)  

x = moles glucose = 0.777 moles  

and  molality = moles solute / kg solvent, molality = 0.777 m

Answer 2

Answer: 7.34 mol/kg.

Explanation: As the relative lowering of vapor pressure is directly proportional to the amount of solute. dissolved.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=x_{solute}$

where,

$p^o$ = vapor pressure of pure solvent  (water) = 760 mmHg

$p_s$ = vapor pressure of solution  = 750 mmHg

$x_{solute}$  = mole fraction of solute

Now put all the given values in this formula ,we get the vapor pressure of the solution.

$\frac{760-750}{760}=x_{solute}$

$x_{sucrose}=0.013$

$x_{water}+x_{sucrose}=1$

$x_{water}=1-0.013=0.987$

A we know :

$\frac{x_{sucrose}}{x_{water}}=\frac{\frac{n_{sucrose}}{n_{sucrose}+n_{water}}}{\frac{n_{water}}{n_{sucrose}+n_{water}}}=\frac{n_{sucrose}}{n_{water}}$

Thus moles of water= 0.987

moles of sucrose = 0.013

Molality : It is defined as the number of moles of solute present per kg of solvent .

Formula used :

$Molality=\frac{n\times 1000}{W_s}$

where,

n= moles of solute  (sucrose) = 0.013

$W_s$ = weight of solvent in g  =$0.0987\times 18=1.77g$

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{0.013moles\times 1000}{1.77g}=7.34mole/kg$

Therefore, the molality of solution will be 7.34 mole/kg.