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
50ml of 20.8% bacl2 and 100ml of 9.8% h2s04 solutions are mixed. Molarity of cl ions in the resulting solution is At. Wt of Ba = 137
Thank all tutors for your help me again
I answered both question earlier. You have a lot of information that is not needed. Changing 980 % to 9.80% makes it make more sense but that number isn't needed either.
To find the final molarity of the H2SO4 solution after dilution, we can use the equation:
M1V1 = M2V2
Where:
M1 = initial molarity of the H2SO4 solution
V1 = initial volume of the H2SO4 solution
M2 = final molarity of the H2SO4 solution
V2 = final volume of the solution after dilution
Given:
M1 = 1 M
V1 = 100 ml
V2 = 500 ml (100 ml H2SO4 solution + 400 ml water)
Density of initial solution = 1.5 g/ml
Density of final solution = 1.25 g/ml
First, let's calculate the initial mass of the H2SO4 solution:
Mass = density × volume
Initial mass = 1.5 g/ml × 100 ml = 150 g
Next, we need to find the final mass of the solution after dilution. Since there is no change in the number of moles of H2SO4 before and after mixing, the mass of the H2SO4 remains the same. Therefore:
Final mass = 150 g
Finally, we can calculate the final molarity using the equation:
M2 = (Final mass / Final volume) × (1 / Molecular weight)
The molecular weight of H2SO4 is:
H = 1 g/mol
S = 32 g/mol
O = 16 g/mol
Molecular weight of H2SO4 = (2 × 1 g/mol) + 32 g/mol + (4 × 16 g/mol) = 98 g/mol
Plugging in the values:
M2 = (150 g / 500 ml) × (1 mol / 98 g)
Converting ml to L:
M2 = (150 g / 0.5 L) × (1 mol / 98 g)
M2 = 300 / 98 mol/L
M2 ≈ 3.06 M (rounded to two decimal places)
Therefore, the final molarity of the H2SO4 solution is approximately 3.06 M.
For the second part of the question:
To calculate the molarity of Cl ions in the resulting solution, we need to find the number of moles of Cl ions in BaCl2 and calculate the final volume of the solution.
Given:
50 ml of 20.8% BaCl2 solution
100 ml of 9.8% H2SO4 solution
Molar mass of Ba = 137 g/mol
First, let's calculate the number of moles of Cl ions in BaCl2:
Moles of Cl ions = (Volume of BaCl2 solution × Molarity of BaCl2) × (Number of Cl ions / Number of BaCl2 ions)
Since BaCl2 dissociates into 2 Cl ions and 1 Ba ion:
Number of Cl ions = 2
Number of BaCl2 ions = 1
Plugging in the values:
Moles of Cl ions = (50 ml × 20.8%) × (2 / 1)
Moles of Cl ions = 10.4 ml × 2 mol/L
Converting ml to L:
Moles of Cl ions = 0.0104 L × 2 mol/L
Moles of Cl ions = 0.0208 mol
Now, let's calculate the final volume of the solution:
Final volume = Volume of BaCl2 solution + Volume of H2SO4 solution
Final volume = 50 ml + 100 ml = 150 ml
Converting ml to L:
Final volume = 150 ml / 1000 ml/L
Final volume = 0.150 L
Finally, we can calculate the molarity of Cl ions using the equation:
Molarity of Cl ions = Moles of Cl ions / Final volume
Plugging in the values:
Molarity of Cl ions = 0.0208 mol / 0.150 L
Molarity of Cl ions ≈ 0.139 M (rounded to three decimal places)
Therefore, the molarity of Cl ions in the resulting solution is approximately 0.139 M.
You're welcome! If you have any more questions, feel free to ask.