Question

a solution of crtric acid C6H8O7 in 50g of acetic acid has a boiling point of elevation of 17.6K.if Kb for acetic acid is 3.07k kg mol, what is the molality of solution?​

Answer 1

Explanation:

Explanation:

Given: 80 gram of H_2SO_4H

2

SO

4

are dissolved in 100 ml of solution.

Thus if 100 ml of solution contain = 80 gram of H_2SO_4H

2

SO

4

1000 ml of solution will contain =\frac{80}{100}\times 1000=800

100

80

×1000=800 gram of H_2SO_4H

2

SO

4

Density of solution= 1.80 g/ml

Now we have to calculate the mass of solution.

Density=\frac{Mass}{Volume}Density=

Volume

Mass

\text {Mass of solution}=Density\times Volume=1.80\times 1000ml=1800gMass of solution=Density×Volume=1.80×1000ml=1800g

Mass of solution = Mass of solute + Mass of solvent

Mass of solvent = mass of solution - mass of solute= (1800- 800)g = 1000 g

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}Molality=

W

s

n×1000

where,

n= moles of solute

Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{800g}{98}=8.2molesMoles=

Molar mass

Given mass

=

98

800g

=8.2moles

W_sW

s

= weight of solvent in g = 1000g

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

Molality=\frac{8.2moles\times 1000}{1000g}=8.2mole/kgMolality=

1000g

8.2moles×1000

=8.2mole/kg