Question

2. What is molarity of 35% (w/v) solution of H2SO4?
(1) 1.25 M
(2) 3.57 M.
(3) 4.23 M
(4) 2.65 M.​

Answer 1

Answer:

Molarity of 35% (w/v) H₂SO₄ = 3.57 M

Explanation:

Molarity is defined as the number of moles per liter of solution.

ie, Molarity = $\frac{no of moles}{volume of solution}$(L)

number of moles = $\frac{given weight}{molecular weight}$

35% (w/v) H₂SO₄, indicates 35g of  H₂SO₄ in 100ml solution

So, in 1000 ml(1L) 350 gm of H₂SO₄ is present.

given weight = 350g

molecular weight = 98g

volume of solution in L = 1

no of moles = $\frac{350g}{98g}$ = 3.57

molarity = $\frac{3.57}{1}$= 3.57 M

Answer 2

Given - Concentration = 35% (w/v)

Find - Molarity

Solution - As per the given concentration of 35% (w/v), there is 35 gram of H2SO4 in 100 ml of solution.

So, concentration of H2SO4 in 1000 ml of solution is 350 gram.

Now, calculating number of moles.

Number of moles = mass/ molar mass

Mass = 350 gram

Molar mass = 98 gram

Number of moles = 350/98

Number of moles = 3.57

Molarity = number of moles/volume of solution in litre

Molarity = 3.57/1

Molarity = 3.57 M

Hence, molarity is (2) 3.57 M.