Question

Calculate the molality of 1 litre solution of 90% H2SO4(weight/volume). The density of the solution is 1.80 g mL-1

Answer 1

we know that
Molarity = [(w/v)% × d × 10]/GMM
GMM= Gram Molar Mass of H2SO4 = 98g
d= density = 1.80g/ml
w/ v =(weight / volume)%

So Molarity(M) = (90×1.80×10)/ 98
M = 16.53M
Hope it Helps
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Answer 2

Answer:

10.2 m

Explanation:

v of solution = 1 l = 1000 ml

D of solution = 1.8 g ml⁻¹

m of solution = D x v

= 1.8 x 1000

= 1800 g

90% H₂SO₄ (w/v) means 90g H₂SO₄ is dissolved in 100 ml of solution.

So , 1000 ml solution will have 900 g H₂SO₄ .

m of solvent = 900 g = 0.9 kg

no. of moles of H₂SO₄ = 900/98 = 9.184

Molality , m = no. of moles / m of solvent in kg

= 9.184/0.9

= 10.2 m

Hope this helps .