Question

molarity of 29%(w/w) H2 S o4 solution whose density is 1.2 gram per ml is:
(a)1.8M
(b)3.6M
(c)2.4M
(d)1.2M​

Answer 1

Answer:3.6M

Explanation:

Use formula of molarity

Answer 2

Molarity of 29% (w/w) H2SO4 solution whose density is 1.2 gram per ml is- b)3.6M.

Given:

The density of 29% (w/w) H2SO4 solution is 1.2 grams per ml.

To find:

The molarity of given H2SO4 solution.

Solution:

Before proceeding to the solution, we should know that the molarity of a solution is equal to the number of moles of the solute divided by the volume of solution in litres. Thus,

$M = \frac{n}{V}$

where n = number of moles = given mass/molar mass and V = volume of solution in litres.

Thus, as given, we have,

29℅ w/w means 29 grams of H2SO4 in 100g of solution.

The molar mass of H2SO4 = 98g (2g+32g+16g×4)

Hence, the number of moles 'n' is

$= \frac{29}{98}$

Now,

The density of solution = mass of solution /volume of solution = m/V

$1.2 = \frac{100}{V}$

$V = \frac{100}{1.2} mL$

To find molarity, the volume of solution should be in litres. So,

$V = \frac{100}{1.2 \times1000 }L$

(as 1 litre = 1000mL)

So,

The molarity of the H2SO4 solution is

$M = \frac{ \frac{29}{98} }{ \frac{100}{1.2 \times 1000} }$

On solving above, we have,

$M = \frac{29 \times 1.2 \times 1000}{98 \times 100}$

$M = \frac{12 \times 29}{98}$

$M = 3.55M$

which is approximately 3.6M.

Hence, the molarity of the 29%(w/w) H2SO4 solution having a density of 1.2 grams per ml is 3.6M.