Question

If m/20 h2so4 solution has 2.5 g/cc density ,what is the molality of solution

Answer 1

Explanation:

Molecular weight of sulphuric acid

=

98

g/mol

There's

20

g

acid in

100

g

of water.

Thus actual mass of water present is

80

g

.

Molality means number of solute particles present per

k

g

of solvent.

Thus,

80

g

of solvent(water) has

20

g

acid

⇒

1

g

of solvent has

20

80

g

acid.

⇒

1000

g

(

=

1

k

g

)

of solvent has

20

80

×

1000

=

250

g

of acid.

Now,

98

g

of acid means

1

mole of acid.

⇒

1

g

of acid means

1

98

moles of acid.

⇒

250

g

of acid means

1

98

×

250

=

2.55

moles of acid.

Thus, molality of the solution is 2.55 molal because 2.55 moles of acid are present in 1000g of water

r

r

r

r

Or going by the conventional method, and applying the formula,

Molality

=

given mass

molar mass

×

1000

mass of solvent in g

=

20

98

×

1000

80

=

2.55

mol/kg

Answer 2

Answer:

0.02 mol/kg is the molality of solution.

Explanation:

Molarity of the solution = M/20 = 0.05 M

0.05 moles of sulfuric acid is present in 1 L of solution.

Mass of 0.05 moles of sulfuric acid = 0.05 mol × 98 g/mol= 4.9 g

Density of the solution = $2.5 g/cm^3=2.5 g/mL$

Mass of the solution :M

Volume of the solution = 1L = 1000 mL

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

Mass of the solution = $2.5 g/mL\times 1000 mL=2500 g$

Mass of the solvent = 2500 g-4.9 g = 2,495.1 g=2.495 kg

Molality =$\frac{moles}{\text{mass of solvent in kg}}$

$molality=\frac{0.05 mol}{2.495 kg}=0.02 mol/kg$

0.02 mol/kg is the molality of solution.