A 55.29-g sample of Ba(OH)2 is dissolved in enough water to make 1.20 L of solution. How many milliliters of this solution must be diluted with water in order to make 1.00 L of 0.100 M Ba(OH)2? (Ignore significant figures for this problem.)
mols Ba(OH)2 = grams/molar mass = approx 0.3 but you need to be more accurate.
M Ba(OH)2 = about 0.3/1.20L = approx 0.31 or so.
Then use c1v1 = c2v2
0.31 x v = 0.100 x 1.00
Solve for v.
512
To solve this problem, we need to determine the volume of the given Ba(OH)2 solution that needs to be diluted to obtain the desired concentration.
Let's start by calculating the number of moles of Ba(OH)2 in the 1.20 L solution using the given mass and molar mass of Ba(OH)2.
1. Calculate the molar mass of Ba(OH)2:
Molar mass (Ba) = 137.33 g/mol
Molar mass (O) = 16.00 g/mol
Molar mass (H) = 1.01 g/mol
Molar mass (Ba(OH)2) = 137.33 g/mol + (2 * 16.00 g/mol) + (2 * 1.01 g/mol) = 171.34 g/mol
2. Calculate the number of moles of Ba(OH)2:
Number of moles = mass / molar mass
Number of moles = 55.29 g / 171.34 g/mol ≈ 0.323 moles
Next, we need to determine the volume of the 1.00 L solution with a concentration of 0.100 M Ba(OH)2.
3. Calculate the number of moles required for the desired concentration:
Molarity (M) = moles / volume (L)
0.100 M = moles / 1.00 L
moles = 0.100 M * 1.00 L = 0.100 moles
Finally, we can calculate the volume of the Ba(OH)2 solution that needs to be diluted:
4. Calculate the volume of the Ba(OH)2 solution to be diluted:
Volume of solution to be diluted = (moles required / moles in original solution) * volume of original solution
Volume of solution to be diluted = (0.100 moles / 0.323 moles) * 1.20 L
Volume of solution to be diluted ≈ 0.310 L or 310 ml
Therefore, approximately 310 mL of the Ba(OH)2 solution must be diluted with water to make 1.00 L of 0.100 M Ba(OH)2.
To find the answer to this question, we need to use the concept of molarity (M) and the equation:
M1V1 = M2V2
where:
M1 = initial molarity of the solution
V1 = initial volume of the solution
M2 = final molarity of the solution
V2 = final volume of the solution
In this problem, we are given:
M1 = unknown
V1 = unknown
M2 = 0.100 M (the desired final molarity of the solution)
V2 = 1.00 L (the desired final volume of the solution)
To solve for the unknowns, let's work through the problem step by step:
1. Calculate the number of moles of Ba(OH)2 in the initial sample:
To do this, we need to use the molar mass of Ba(OH)2, which is:
Molar mass of Ba(OH)2 = 137.33 g/mol (1 Ba x 137.33 g/mol) + (2 O x 16.00 g/mol) + (2 H x 1.01 g/mol)
= 137.33 g/mol + 32.00 g/mol + 2.02 g/mol
= 171.35 g/mol
Number of moles of Ba(OH)2 = mass of sample (g) / molar mass (g/mol)
Number of moles of Ba(OH)2 = 55.29 g / 171.35 g/mol
2. Calculate the initial volume of the solution:
To do this, we need to use the initial molarity (unknown) and the number of moles of Ba(OH)2 from step 1.
M1V1 = moles of Ba(OH)2
V1 = moles of Ba(OH)2 / M1
V1 = (55.29 g / 171.35 g/mol) / M1
3. Solve for M1:
Rearrange the equation from step 2 to solve for M1.
M1 = (55.29 g / 171.35 g/mol) / V1
4. Calculate the volume of the solution that needs to be diluted:
Using the equation M1V1 = M2V2 and the values we know (M1, V1, M2, and V2), we can solve for V1.
M1V1 = M2V2
V1 = (M2V2) / M1
V1 = (0.100 M x 1.00 L) / M1
Finally, we have the equation for V1, which is the volume of the solution that needs to be diluted to make 1.00 L of 0.100 M Ba(OH)2.