Question

calculate the number of molecules present in 2.5 Litre of sulphuric acid having density 1.21 gm/l​

Answer 1

Answer:

Given : Molarity of H 2 SO 4

= 18M

As, Molarity = volume of solution

no. of moles of solute

i.e., 18moles of sulphuric acid is present per litre of solution.

Density=1.8g/mL {given}

Thus , mass of one litre of this solution ,

Density = volumemass

mass = density×volume

mass = 1.8×1000

mass = 1800g

Thus , mass of 1litre 18M H 2 SO 4

solution is 1800g or 1.8kg.

Now, mass of solution = mass of solute + mass of solvent

mass of solution = no. of moles of solute X molar mass of solute + mass of solvent

1800 = 18×98 + mass of solvent

mass of solvent = (1800−1764)g

mass of solvent = 36g

Now , molality of H 2 SO 4

solution is given by,

molality =

massofsolvent(inkg)

numberofmolesofsolute

molality = 38

18×100

molality = 500m

Hence, the molality of H 2SO4

solution is 500m.

Explanation:

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