Question

What mass in grams of glucose (MM = 180.2 g/mol) would need to be dissolved in 100.0 g of water to decrease the vapor pressure of the solution to 22.0 torr? (P° of water is 23.5 torr)

Answer 1

Answer:

may be my answer helps you

Explanation:

water 20 , the vapour pressure of the resulting solution will be;

No. of moles of glucose=

180

18

=0.1

No. of moles of water=

18

178.2

=9.9

Mole fraction of glucose=

4.9+0.1

0.1

=

10

0.1

=0.01

Using Raoult's law, as solute is non-volatile,

x

β

=

P

A

o

P

A

o

−P

A

P

A

o

=vapour pressure of pure water

P

A

=vapour pressure of the solution

x

β

=mole fraction of glucose

⟹0.01=

17.5

17.5−P

A

⟹P

A

=17.5−0.175=17.325

The vapor pressure of the mixture=17.325 mm Hg.

Hence, the correct option is D

Answer 2

Given: Mass of water = 100 g

           Vapour Pressure of Water = 23.5 torr

           Final Vapour Pressure = 22.0 torr

To Find : Mass of glucose to be added to solution

Solution:

• For solving this question, we will use the colligative property- relative lowering of vapour pressure (RLVP)

                 RLVP ⇒ $\frac{P_{0} - P_{s} }{P_{0} }$ = $\frac{x_{glu} }{x_{glu}+ x_{water} }$                (1)

where, $P_{0}$ is vapour pressure of water = 23.5 torr

          $P_{s}$ is vapour pressure of solution = 22 torr

          $x_{glu}$ is no. of moles of glucose

          $x_{water}$ is no. of moles of water = $\frac{100}{18}$

Substituting these values in (1):

                    $\frac{23.5 - 22}{23.5}$ = $\frac{x_{glu} }{x_{glu} + \frac{100}{18} }$

                   ⇒ $\frac{1.5}{23.5}$ = $\frac{x_{glu} }{x_{glu} + 5.5 }$                      

                   ⇒  3($x_{glu}$ + 5.5) = 47 $x_{glu}$                  (reducing LHS to $\frac{3}{47}$)

                   ⇒ 3$x_{glu}$ + 16.5 = 47 $x_{glu}$

                   ⇒ 44 $x_{glu}$ = 16.5

                   ⇒ $x_{glu}$ = 0.375

   Mass of glucose = no. of moles × molar mass

   Mass of glucose = 0.375 × 180.2

  Mass of glucose = 67.575 g

Therefore, 67.575 g of glucose is required to decrease the vapour pressure of the solution.