describe the preparation of 100ml of 6.0M HCl from a concentrated solution that has a specific gravity of 1.18 and is 37% w/w HCl.
To prepare 100ml of 6.0M HCl from a concentrated solution with a specific gravity of 1.18 and is 37% w/w HCl, you will need to use some simple calculations based on the concentration and volume of the concentrated solution.
Here's how you can go about it:
1. Determine the amount of HCl in the concentrated solution:
- The specific gravity of the solution is 1.18, which means that it is 1.18 times denser than water.
- The weight percent concentration (w/w) of HCl is 37%, which means that for every 100g of the solution, 37g is HCl.
- Therefore, the weight of HCl in 100g of the solution is (37g/100g) * 100g = 37g.
2. Convert the weight of HCl to its molar amount:
- Find the molar mass of HCl, which is approximately 36.46 g/mol.
- Divide the weight of HCl by its molar mass: 37g / 36.46 g/mol ≈ 1.014 mol.
3. Determine the volume of the concentrated solution needed:
- Concentration is defined as moles of solute per liter of solution (Molarity = moles/volume).
- You want to prepare a 100ml solution with a concentration of 6.0M.
- Rearranging the equation gives: volume = moles / concentration.
- So, the volume of the concentrated solution needed is: 1.014 mol / 6.0M ≈ 0.169L (or 169ml).
4. Dilution process:
- Take the 169ml of the concentrated HCl solution and transfer it into a suitable container.
- Add distilled water slowly and carefully to reach a final volume of 100ml, while stirring to ensure thorough mixing.
- Confirm the final volume is exactly 100ml.
By following these steps, you should be able to prepare 100ml of 6.0M HCl from the given concentrated HCl solution.
Convert 37% HCl to M.
1.l8 g/mL x 1000mL x 0.37 x (1/36.5 = ?M.
Then c1v1 = c2v2
c = concn
v = volume.
12%
To prepare 100ml of 6.0M HCl from a concentrated solution with a specific gravity of 1.18 and is 37% w/w HCl, follow these steps:
Step 1: Determine the amount of concentrated solution needed
To calculate the volume of the concentrated solution needed, we use the formula:
Volume (V1) x Concentration (C1) = Volume (V2) x Concentration (C2)
Let V1 be the volume of concentrated solution needed (in ml),
C1 be the concentration of concentrated solution (37% w/w HCl),
V2 be the final volume required (100 ml),
C2 be the final concentration required (6.0M).
Substituting the values, we get:
V1 x 0.37 = 100 x 6.0
V1 = (100 x 6.0) / 0.37
V1 ≈ 162.16 ml (rounded to 2 decimal places)
Therefore, you will need approximately 162.16 ml of the concentrated solution.
Step 2: Consider the specific gravity conversion
Since the specific gravity of the concentrated solution is 1.18, we need to convert the volume in milliliters (ml) to grams (g) to account for the specific gravity.
To do this, multiply the volume of the concentrated solution (V1) by its specific gravity (1.18):
162.16 ml x 1.18 = 191.36 grams (rounded to 2 decimal places).
Step 3: Calculate the amount of HCl in the concentrated solution
The concentrated solution is 37% w/w HCl, which means that 37 grams of HCl is present in every 100 grams of the solution.
To calculate the amount of HCl in the concentrated solution:
(37g HCl / 100g solution) x 191.36g solution = 70.71g HCl (rounded to 2 decimal places)
Step 4: Dilution
Now, we need to dilute the calculated amount of HCl to 100 ml to achieve the desired concentration.
Since the molar mass of HCl is approximately 36.46 g/mol:
Moles of HCl needed = Mass of HCl / Molar mass of HCl
= 70.71g HCl / 36.46 g/mol
≈ 1.94 mol (rounded to 2 decimal places)
To dilute the HCl to 100ml:
Volume (V2) = Moles / Concentration
= 1.94 mol / 6.0 M
≈ 0.32 L or 320 ml (rounded to 2 decimal places)
Step 5: Add distilled water
To prepare 100ml of 6.0M HCl, add enough distilled water to make the total volume 100 ml.
Subtract the volume of HCl from the total volume:
100 ml - 320 ml ≈ -220 ml (negative value obtained because there is excess HCl)
Therefore, you need to add approximately 220 ml of distilled water.
NOTE: It is crucial to perform these calculations accurately and in a well-ventilated area while following proper safety protocols. HCl is a strong acid and can be hazardous if mishandled.