Question

A 0.500 molar (0.500 gmol per litre) solution of sulphuric acid in water flows into a process unit at a rate of 1.25 m3.min-1. The density of the solution is 1.03 kg/litre.

Calculate:
(i) the mass concentration of H2SO4 in kg m-3
(ii) the mass flow rate of H2SO4 (solution) in kg/s
(iii) the mass fraction of H2SO4
Using answer from (i):

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

(i) The mass concentration of H₂SO₄ is 49 Kg/m³

(ii) The mass flow rate of H₂SO₄ (solution) is 1.02 Kg/s

(iii) The mass fraction of H₂SO₄ is 0.0475.

Explanation:

Given, the concentration of Sulphuric acid solution, C = 0.5 M

The rate of flow, f = 1.25 m³/min

The density of solution, ρ = 1.03 Kg/L = 1030 Kg/m³

We know that molar mass of sulphuric acid = 98g/mol

(i) To calculate the mass concentration of H₂SO₄ in Kg/m³ :-

Mass concentration =  C × molar mass of H₂SO₄

$= \frac{0.5mol}{1L}\times\; 98\; g/mol$

$=\frac{49}{10^{-3}m^3}\times 10^{-3}Kg$

$= 49\; Kg/m^3$

(ii)  The mass flow rate of H₂SO₄ (solution) in kg/s:-

Mass flow rate = Flow rate × mass concentration

Mass flow rate $=\frac{ 1.25\;m^3}{1\;min}\times \frac{49Kg}{m^3}= \frac{ 1.25\times 49\;Kg}{60sec} = 1.02 \;Kg/s$

(iii) The mass fraction of H₂SO₄:-

Mass flow rate of the solution = Density × flow rate

$m_{solution}=\frac{1030Kg}{1m^3}\times \frac{1.25m^3}{1\;min}$

$m_{sol}=21.45\; Kg/s$

Mole fraction of H₂SO₄ $=\frac{m_{H_2SO_4}}{m_{sol}}$

Mole fraction of H₂SO₄ $=\frac{1.02Kg/s}{21.45Kg/s}=0.0475$

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