The molarity of a solution of sulphuric acid is 1.35 M. Calculate its

1.	The molarity of a solution of sulphuric acid is 1.35 M. Calculate its molarity (The density of the acid solution is 1.02 g cm3)
Answer» Let the solution be 1 litre or 1000 cm³ ∴ Number of moles of H₂SO₄ = 1.35 Wt. of solution = 1000 x 1.02 = 1020 g Wt. of sulphuric acid = 1.35 x 98 = 132.3g Wt. of water = 1020 - 132.3 = 887.79 Molality of H₂SO₄ = ppp x 1000 = 1.52 m From (i) M₁ = 110.82, from (ii) M₂ = 196.15 AB₄ – AB, = B₂ 196.15 - 110.82 = B₂ 85.33 = B₂ B = 42.665 Molar mass of AB₂ = Atomic mass of A+ x 2 atomic mass of B 110.82 = Atomic mass of A + 85.33 Atomic mass of A = 110.82 - 85.33 = 25.49 Atomic mass of A = 25.499 Atomic mass of B = 42.669

The molarity of a solution of sulphuric acid is 1.35 M. Calculate its molarity (The density of the acid solution is 1.02 g cm3)

Answer»

Let the solution be 1 litre or 1000 cm³

∴ Number of moles of H₂SO₄ = 1.35

Wt. of solution = 1000 x 1.02 = 1020 g

Wt. of sulphuric acid = 1.35 x 98 = 132.3g

Wt. of water = 1020 - 132.3 = 887.79

Molality of H₂SO₄ = ppp x 1000 = 1.52 m

From (i) M₁ = 110.82, from (ii) M₂ = 196.15 AB₄ – AB, = B₂

196.15 - 110.82 = B₂ 85.33 = B₂ B = 42.665

Molar mass of AB₂ = Atomic mass of A+ x 2 atomic mass of B

110.82 = Atomic mass of A + 85.33 Atomic mass of A = 110.82 - 85.33 = 25.49

Atomic mass of A = 25.499

Atomic mass of B = 42.669

Discussion

No Comment Found

Related InterviewSolutions

Define Solutions.
At `27^(@)C`,3.92 gm `H_(2)SO_(4)` is present in 250 ml solution. The osmotic pressure of this solution is 1.5 atm. If the osmotic pressure of solution of NaOH is 2 atm at same temperature, then concentration of NaOH solution isA. 0.32 MB. 12.183 MC. `72.3 gm//lit`D. 1 M `Na_(2)SO_(4)`
If K_(3)[Fe(CN)_(6)]` is completely ionised in a solution, what is the number of paritcles forms when one molecule is completely ionised.
29.2% (W/W) HCI stock solution has a density of 1.25 g `mL^(-1)`. The molecular mass of HCI is 36.5 g `mol^(-1)` Calculate the volume (mL) of stock solution required to prepare 200 mL of 0.4 M HCI
A compound `H_(2)X` with molar mass of `80 g` is dissolved in a solvent having density of `0.4 g mL^(-1)`. Assuming no change in volume upon dissolution, the molality of a `3.2` molar solution isA. `9`B. `6`C. `5`D. `8`
A compound `MX_(2)` with molar mass of 80 g is dissolved in a solvent having density 0.4 g `mL^(-1)`. Assuming that there is no change in volume upon dissociation, what is the molality of a 3.2 molar solution ?
The solubility of a solute in a solvent (that is, the extent of the mixing of the solute and solvent species) depends onA. the natural tendency for the solute and solvent species to mixB. the natural tendency for a system to have the lowest energy possibleC. a balance between (1) and (2)D. neither (1) and (2)
Maximum amount of a solid aolute that can be dissolved in a specified amount of a given liquid solvent does not depend upon………..A. temperatureB. nature of soluteC. pressureD. nature of solvent
Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent. It depends upon (i)nature of solute , (ii)nature of solvent , (iii)temperature , (iv)pressureA. only (i),(ii) and (iii)B. only (i), (iii) and (iv)C. only (i) and (iv)D. (i),(ii),(iii) and (iv)
A complex is represented as `CoCl_(3) . XNH_(3)`. Its `0.1` molal solution in aqueous solution shows `Delta T_(f) = 0.558^(circ). (K_(f)` for `H_(2)O` is `1.86 K "molality"^(-1))` Assuming `100%` ionisation of complex and co-ordination number of `Co` as six, calculate formula of complex.