A 25.00mL sample of a household cleaning solution was diluted to 250.0mL in a volumetric flask. A 50.00mL aliquot of this solution required 40.38mL of 0.2506 M HCl to reach a bromocresol green end point. Calculate the weight/volume percentage of NH3 in the sample. (Assume that all the alkalinity results from the ammonia.)

Can someone please explain how to get the volume when calculating for the weight/volume percentage. I know how to get the weight but don't know how to get the volume for this question. Please explain it throughly. Thanks.

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1mmolNH17.031gNH0.1943mmolHCl× 41.27mLHCl × ×mL1mmolHCl1000mmol × 100%50.00mL × 25.00mL250.0mL = 2.731% (w/v)NH

NH3 + HCl = NH4Cl

[(0.1943mmol/mL HCl)× 41.27mL HCl × (1mmol NH3/ 1mmol HCl) x 17.031g NH3 /1000mmol NH3)] / [50.00mL × 25.00mL250.0mL ] x100% = 2.731% (w/v)NH3

Well, calculating the weight/volume percentage involves finding the weight of the solute (in this case, NH3) and the volume of the solvent (in this case, the cleaning solution) in the sample.

To find the weight of NH3, you need to consider the reaction between NH3 and HCl. The balanced equation is:

NH3 + HCl -> NH4Cl

From the equation, you can see that 1 mole of NH3 reacts with 1 mole of HCl to form 1 mole of NH4Cl. So, the moles of NH3 in the aliquot can be calculated using the volume and molarity of HCl used:

moles of NH3 = moles of HCl

Now, to find the moles of HCl, you can use the equation:

moles of HCl = concentration of HCl x volume of HCl (in liters)

Converting the volume of HCl from mL to liters (40.38 mL = 0.04038 L) and plugging in the values, you can calculate the moles of NH3.

Once you have the moles of NH3, you can find the weight of NH3 (in grams) using the molar mass of NH3, which is approximately 17.03 g/mol.

Now, to calculate the volume of the cleaning solution, you need to use the dilution formula:

C1V1 = C2V2

where C1 and V1 are the concentration and volume of the aliquot, and C2 and V2 are the concentration and volume of the diluted solution.

Given that the concentration of the aliquot is 0.2506 M, the volume of the aliquot is 50.00 mL, and the volume of the diluted solution is 250.0 mL, you can plug in these values to find the concentration of the diluted solution.

Finally, multiplying the concentration of the diluted solution by its volume, you can determine the weight of the cleaning solution.

To find the weight/volume percentage, divide the weight of NH3 by the weight of the cleaning solution and multiply by 100. This will give you the weight/volume percentage of NH3 in the sample.

I hope this explanation helps! If you have any more questions, feel free to ask.

To calculate the weight/volume percentage of NH3 in the sample, you need to determine the volume of the solution containing NH3. Here's how you can calculate it:

1. Start by calculating the concentration (molarity) of the HCl solution used to reach the bromocresol green end point. The given concentration is 0.2506 M.

2. Use the volume and concentration of the HCl solution to calculate the number of moles (n) of HCl used in the titration. The equation you can use is:

n = C * V

Where:
n is the number of moles
C is the concentration of the HCl solution in M (0.2506 M in this case)
V is the volume of the HCl solution used in the titration (40.38 mL in this case)

3. Since HCl and NH3 react in a 1:1 ratio, the number of moles of HCl used in the titration is equal to the number of moles of NH3 present in the sample.

4. Now, you need to calculate the number of moles of NH3 in the 50.00 mL aliquot. To do this, use the following equation:

mols NH3 = n/V_aliquot

Where:
mols NH3 is the number of moles of NH3 in the aliquot
n is the number of moles of NH3 in the sample (calculated in step 3)
V_aliquot is the volume of the aliquot used in the titration (50.00 mL in this case)

5. Next, calculate the number of moles of NH3 in the entire 250.0 mL solution. Since you diluted the 25.00 mL sample to 250.0 mL, you can use the following equation:

mols NH3_total = (mols NH3) * (250.0 mL / 25.00 mL)

Where:
mols NH3_total is the number of moles of NH3 in the entire 250.0 mL solution
mols NH3 is the number of moles of NH3 in the aliquot (calculated in step 4)

6. Finally, use the weight and volume of the sample to calculate the weight/volume percentage of NH3. The equation for weight/volume percentage is:

Weight/volume percentage = (Weight of NH3 / Volume of solution) * 100

Where:
Weight of NH3 is the molar mass of NH3 multiplied by the number of moles of NH3 (calculated in step 5)
Volume of solution is 250.0 mL

By following these steps, you should be able to calculate the weight/volume percentage of NH3 in the sample.

You took a 25.00 mL sample and diluted t 250.0 mL and titrated a 50.00 mL aliquot. You have the weight in the 50.00 mL sample. The weight in the original 25.00 mL sample is weight in titrated sample x (250/50) = ? Now you have the weight in the original 35.00 mL sample and that is the volume you use.