100ml of h2s04 solution having molarity 1M and density 1.5g/ml is mixed with 400ml of water. calculate final molarity of h2s04 solution, if final density is 1.25g/ml.

Question

100ml of h2s04 solution having molarity 1M and density 1.5g/ml is mixed with 400ml of water. calculate final molarity of h2s04 solution, if final density is 1.25g/ml.

Answer 0.227M

Who helps me to solve it

Thank you a lot

Answer 1

To calculate the final molarity of the H2SO4 solution, we need to use the equation:

M1V1 + M2V2 = M3V3

Where:
M1 = initial molarity of the H2SO4 solution
V1 = initial volume of the H2SO4 solution
M2 = molarity of water (0 since water does not have a molarity)
V2 = volume of water added
M3 = final molarity of the H2SO4 solution
V3 = final volume of the diluted solution

Given:
M1 = 1M
V1 = 100ml
M2 = 0 (since water does not have a molarity)
V2 = 400ml
M3 = ?
V3 = V1 + V2 = 100ml + 400ml = 500ml

Now, let's substitute the given values into the equation and solve for M3:

(1M)(100ml) + (0)(400ml) = M3(500ml)
100ml + 0 = M3(500ml)
100ml = 500M3
M3 = 100ml / 500ml
M3 = 0.2M

So, the final molarity of the H2SO4 solution is 0.2M.

However, the question also asks to calculate the final density of the solution. To do that, we can use the equation:

Density = (mass of the solution) / (volume of the solution)

Given:
Initial density = 1.5g/ml
Final density = 1.25g/ml
Volume of the solution = 500ml

Let's calculate the mass of the solution:

Mass of the solution = density × volume of the solution
Initial mass = (1.5g/ml) × 500ml = 750g
Final mass = (1.25g/ml) × 500ml = 625g

Therefore, the mass of H2SO4 before dilution = Initial mass - Final mass
Mass of H2SO4 = 750g - 625g = 125g

Now, let's calculate the new molarity of the H2SO4 solution:

Molarity = (moles of solute) / (liters of solvent)

First, let's calculate the moles of H2SO4:

Moles of H2SO4 = Mass of H2SO4 / Molar mass of H2SO4

The molar mass of H2SO4 is:
(2 × atomic mass of H) + atomic mass of S + (4 × atomic mass of O)
(2 × 1.008g/mol) + 32.06g/mol + (4 × 16.00g/mol)
2.016g/mol + 32.06g/mol + 64.00g/mol
= 98.086g/mol

Moles of H2SO4 = 125g / 98.086g/mol
Moles of H2SO4 = 1.273 mol

Now, let's calculate the final volume of the solution:

Final volume of the solution = Volume of the solution = 500ml

Finally, let's calculate the new molarity of the H2SO4 solution:

Molarity = Moles of H2SO4 / Final volume of the solution
Molarity = 1.273 mol / (500mL / 1000)
Molarity = 1.273 mol / 0.5 L
Molarity = 2.546 mol/L

Approximating to three significant figures, the final molarity of the H2SO4 solution is 2.55M.