To prepare 1L of 75% H2SO4 from concentrated H2SO4 (98% of density 1.84g/ml), you can use the given equation, but first, let me explain how it works.
In this equation, V represents the volume of concentrated H2SO4 needed to prepare 1L of 75% H2SO4 solution. The equation is based on the principle of conservation of mass, where the mass of the concentrated H2SO4 should be equal to the mass of the diluted solution.
Now, let's break down the equation step by step:
1. Vx1.84x0.98:
Here, V multiplied by 1.84 gives the mass of concentrated H2SO4 required. The density of 1.84 g/ml is used to convert the volume into mass. Then, multiplying by 0.98 accounts for the purity of the concentrated H2SO4 (98%).
2. (1000-V)+ 1.84V:
This expression calculates the mass of the diluted solution. (1000-V) represents the mass of the solvent (water) needed, and 1.84V represents the mass of the concentrated H2SO4 being mixed.
3. (1000-V)+ 1.84V)0.75:
Finally, multiplying the total mass of the diluted solution by 0.75 gives you the mass of the final solution (75% H2SO4), as a percentage of the total mass.
By equating the two expressions, you ensure that the mass of the concentrated H2SO4 is equal to the mass of the diluted solution. Therefore, when you solve this equation, you will find the value of V, representing the volume of concentrated H2SO4 needed.
Now, let's solve the equation step by step:
Vx1.84x0.98 = (1000-V) + 1.84V)0.75
1.84V x 0.98 = (1000 - V) + 1.84V)0.75
1.7992V = 1000 - V + 1.38V
1.7992V = 1000 + 0.38V
1.4192V = 1000
V = 1000/1.4192
V ≈ 703.2 ml
So, approximately 703.2 ml of concentrated H2SO4 is needed to prepare 1L of 75% H2SO4 solution.