add 2.4ml glacial aetic acid in one liter of d.m water
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To prepare a 5000 PPM (parts per million) acetic acid solution, you would need to dissolve a specific amount of acetic acid (in grams) in a known volume of water (in liters). The formula to calculate the amount of acetic acid needed is: Mass of acetic acid (g) = volume of solution (L) x desired concentration (PPM) / 1000000.
To prepare a 10M solution of acetic acid, dissolve 60.05g of glacial acetic acid (CH3COOH) in enough water to make a final volume of 1 liter. The molar mass of acetic acid is 60.05 g/mol. Make sure to wear appropriate safety gear, as acetic acid is corrosive.
To prepare the 0.50M acetic acid solution, you can use the formula C1V1 = C2V2. Plugging in the values, you get (2.5M)(V1) = (0.50M)(100.0mL). Solving for V1 gives V1 = 20.0 mL. Therefore, 20.0 milliliters of the 2.5M stock solution is required to prepare 100.0 milliliters of the 0.50M acetic acid solution.
To prepare a 0.83N acetic acid solution, you would mix 1 volume of glacial acetic acid (100% purity) with 9.1 volumes of water. This solution will have a concentration of 0.83N. Remember to handle glacial acetic acid safely as it is corrosive.
To prepare a 0.1 M solution of acetic acid in 100 ml, you would need to dissolve 1.04 grams of acetic acid in enough water to make 100 ml of solution. The molecular weight of acetic acid is 60.05 g/mol. So, 0.1 M solution is achieved by using the formula: (0.1 mol/L) x (60.05 g/mol) = x g/L. Thus, in 100 ml, you would need 1.04 grams of acetic acid.
To prepare a solution containing 3% acetic acid and 97% water, you can add the required amount of acetic acid to the water based on the desired volume of the final solution. For example, to make 100mL of this solution, you would add 3mL of acetic acid to 97mL of water. Make sure to mix thoroughly to ensure homogeneity of the solution.