Applying the law of conservation of mass, which suggests that the moles of Ethanol in x ml of 95% ethanol solution is equal to the moles of ethanol in 125 ml of 75% ethanol solution.
$$\implies \mathrm{M_1V_1 = M_2V_2}$$
Here, in this case: $$\mathrm{M_1 = 0.95; \ V_1 = x\ ml; \ M_2 = 0.75; \ V_2=125\ ml}$$
$$\implies \mathrm{V_1 = 98.684 \ ml}$$
But the final volume $$[\text V_2]$$ = 125 ml
So, for preparing 125 ml of 75% ethanol solution, one must take 98.684 ml of 95% ethanol solution and 26.316 ml of Water.
Preparation of 70% alcohol: 70% alcohol means 70 mL of alcohol should be present in 100 mL of solution. Therefore, if we take 74 mL of the 95% alcohol and make its volume up to 100 mL, the concentration of the obtained solution will be 70% by volume.
for 200 ml you need 140ml of pure ethanol, or 147.4ml of 95% for starters and dilute to 200 ml with water. Take this volume of 95% solution and dilute till final volume of 200 mL with water.
Therefore when we pipette out exactly 8.42- ml into a 10- ml standard flask and then dilute with distilled water up to the mark you will get 80% Ethanol in water. ( ...
Add (10–8.42)= 1.58- ml of distilled water to 8.42- ml of 95% Ethanol and mix, you will get 80% Ethanol.
You take x ml of 70 alcohol. It contains 0.7x ml of pure alcohol and 0.3x ml of water. To make it 50% alcohol you need to have equal amount of water and alcohol in the solution. So, add 0.4x ml of water.
To prepare 20% V/V ethanol. If you can have a 100% V/V ethanol concentration. Measure 20 mL ethanol using graduated cylinder.Placed it in a beaker then add 80 mL water. to make a 100 mL of 20% ethanol.
- Since 1 liter = 1000 mL, we can convert 0.05 liters to mL: 0.05 liters x 1000 mL/liter = 50 mL. Therefore, we need to add 50 mL of water to 950 mL of 95% ethyl alcohol to make 1 liter of a 30% ethyl alcohol solution.
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