How many grams of AgNO3 are needed to make 250 ml of a solution that is 0 145 m?

In order to calculate the amount of silver nitrate (AgNO3) needed to make 250 ml of a solution that is 0. 145 m, you must use a formula to convert molarity (M) to moles (mol). The formula for this conversion is: Molarity (M) = moles/Liters (M=mol/L).

Therefore, you must find the amount of moles that are needed to make a solution of 0. 145 m.

Once you have calculated the moles required, you can use the molar mass of AgNO3 (169. 87 g/mol) to determine the grams of silver nitrate needed. Using the formula M= mol/L, if the molarity of solutions is 0.

145, then the number of moles of silver nitrate needed for 250 ml of solution can be calculated as follows: 0. 145 mol/L x 0. 250 L = 0. 03625 mol.

Finally, to determine the number of grams of AgNO3 required to make a 0. 145 m solution of 250 ml, you must multiply the molar mass of AgNO3 (169. 87 g/mol) by 0. 03625 mol. This calculation yields 6.

12 grams of AgNO3 as the amount of silver nitrate needed to make 250 ml of a 0. 145 m solution.

How many grams are in 250ml of a 0.150 M solution of AgNO3?

By using the equation C = n/V (concentration = moles of solute over volume of solution) we can calculate that for a 0. 150 M solution of AgNO3, the number of moles of solute given 250ml of the solution is 0.

0375 mol. Therefore, since AgNO3 has a molar mass of 169. 87 g/mol, the total number of grams in 250ml of the 0. 150 M solution of AgNO3 is 28. 48 grams.

How much AgNO3 is required to prepare 250ml of 0.05 N aqueous solution?

To prepare a 0. 05 N aqueous solution of AgNO3, you would need to dissolve 0. 125 grams of AgNO3 (silver nitrate) in 250 ml of distilled water. First, weigh the silver nitrate on an analytical or sensitive balance.

Take an appropriately-sized volumetric flask and weigh it so it can be tared (subtracted from the total weight) when done. Then, add the weighed silver nitrate to the flask, fill it up with distilled water, and swirl to dissolve completely.

After the AgNO3 is completely dissolved, mark the flask with the final volume and density. The final solution will be 0. 05 N AgNO3 with a molarity of 0. 05 M.

How many grams is 250 in ml?

250 milliliters (ml) is equal to 250 grams (g). In the metric system, one milliliter is equal to one gram. Therefore, 250 milliliters is equal to 250 grams. In other systems of measurement, there may be slight variations.

For example, in the US Customary System, one milliliter is equal to 0. 033814 ounces, so 250 milliliters is equal to 8. 4535 ounces.

How will you prepare 250ml of 0.05 M acid solution?

To prepare 250 mL of 0.05M acid solution:

1. Measure out 25.5g of the acid of choice in a beaker or other container.

2. Dissolve the acid completely in ~200mL of deionized or distilled water.

3. Pour the solution into a volumetric flask, add water to make up the total volume to the 250mL mark, and mix the solution by shaking and inverting the flask several times.

4. Double-check that the solution has reached the 250mL mark, then store the flask in a safe place.

5. Label the flask with the date, molarity, and type of acid and store in a cool, dark place.

How do you convert 250 mL to grams?

To convert 250 mL to grams, you need to know the density of the substance you’re measuring. Once you know the density, you can use the following formula to calculate the mass in grams: mass (g) = volume (mL) x density (g/mL).

For example, if the density of the substance is 0. 7 g/mL, the calculation would be: mass (g) = 250 mL x 0. 7 g/mL = 175 g.

How do you make 250ml of 0.1 M HCl?

To make 250ml of 0. 1M HCl, you will need 25 grams of hydrochloric acid. First, put on protective gear and safety glasses as hydrochloric acid is a corrosive substance. Weigh out the 25 grams of hydrochloric acid and add it to 250ml of distilled water in a beaker.

Stir until the hydrochloric acid is completely dissolved in the water and the solution is clear. Now measure the solution’s pH with a pH paper and check the result. If it is the same as 0. 1M HCl, then the solution is complete.

If not, then you will need to add more hydrochloric acid or water until you reach the desired pH. Finally, transfer the solution to a different container and store it in a cool and dry place.

How do you prepare and Standardise 250 ml of 0.1 molar Sulphuric acid solution?

To prepare and standardise 250 ml of 0. 1 molar Sulphuric acid solution, you will first need to weigh out 1. 092 g of sulphuric acid (H2SO4) and accurately transfer to a 250 ml volumetric flask. Then add distilled water to the volumetric flask until the mark is reached and mix until the sulphuric acid is dissolved.

You will then need to calibrate a burette with the sulphuric acid solution and accurately dispense exactly 10 ml of the solution into a conical flask. You will then need to prepare a standard solution of Sodium Carbonate (Na2CO3) by accurately weighing out 1.

8065 g of the solid and transferring it to a 250 ml volumetric flask into which you then add distilled water and mix well until the Na2CO3 is completely dissolved. After that, you will need to calibrate a second burette with this standard Na2CO3 solution and slowly add it drop by drop to the acid in the conical flask until you first see a permanent pale pink colour.

This colour change indicates that the acid has been neutralised by the base. After that, you will need to carefully note down the amount of Na2CO3 solution used and calculate the molarity of the sulphuric acid.

The molarity of the sulphuric acid will be equal to the molarity of the Na2CO3 divided by 10. Finally, if the calculated molarity matches the desired molarity, the solution is ready. If not, you should repeat the entire standardisation process until the desired molarity is achieved.

How would you prepare 0.5 M HCl solution in 100 ml?

To prepare 0.5 M HCl solution in 100 ml, you’ll need to find out the mass of HCl that needs to be added to make a 0.5 M solution. To do this you can use the equation:

Mass of solute (g) = Concentration (M) x Volume (L) x Molar Mass (g/mol)

In this equation the concentration is 0.5 M, the volume is 0.1 L and the molar mass of Hydrogen chloride (HCl) is 36.46 g/mol. Therefore, the mass of HCl needed is 0.5 x 0.1 x 36.46 = 1.82 g.

Now you can add 1. 82 g of HCl to a beaker containing 80 ml of deionized water and mix thoroughly to dissolve the HCl. Once the HCl is completely dissolved, you can add more deionized water to make the total volume 100 ml.

Make sure to mix again until you get a homogeneous solution.

Your 0.5 M HCl solution in 100 ml is now ready for use.

How do you make 0.5 N HCl in 1000 ml water?

To make a 0. 5 N HCl solution in 1000 ml of water, you need 260. 25 g of hydrochloric acid and 1000 ml of water. To make the solution: first, measure out 260. 25 grams of hydrochloric acid and place it in an appropriately-sized container.

Next, add 1000 ml of distilled water to the container. Securely close the container and gently mix the contents until all of the acid has dissolved. Make sure to wear proper protective gloves, clothing, and eye protection throughout the entire process.

Once the solution is fully mixed, the 0. 5 N HCl solution can be used as needed or stored in a sealed container for later use.

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