0,98
The mole fraction and molality of ethanol -C2H5OH in an aqueous solution that is 45.0 percent ethanol by volume and the density of water is 1.00g per mL that of ethanol is 0.789 grams per mL and 70/18. A mole fraction in chemistry is the amount that is divided by the total amount of all constituents.
The total volume of the solution is 10ml + 40ml = 50ml. The percentage concentration of ethanol in the solution is (10ml / 50ml) * 100% = 20%.
1. Extract 959,6 mL from the 99 % solution. 2. Add 40,4 mL water.
The mixture water-ethanol is homogeneous.
The polarity difference between ethanol and water affects their interactions in a solution because water is a polar molecule with positive and negative charges, while ethanol is also polar but less so. This difference in polarity allows ethanol and water to mix well together, forming a homogeneous solution. The positive and negative charges in water attract the ethanol molecules, leading to strong interactions between the two substances.
The mole fraction and molality of ethanol -C2H5OH in an aqueous solution that is 45.0 percent ethanol by volume and the density of water is 1.00g per mL that of ethanol is 0.789 grams per mL and 70/18. A mole fraction in chemistry is the amount that is divided by the total amount of all constituents.
To find the mole fraction of ethanol, you first calculate the total moles of the solution, which is 3.00 + 5.00 = 8.00 moles. Then, you divide the moles of ethanol by the total moles of the solution: 3.00 moles / 8.00 moles = 0.375. So, the mole fraction of ethanol in the solution is 0.375.
The total volume of the solution is 10ml + 40ml = 50ml. The percentage concentration of ethanol in the solution is (10ml / 50ml) * 100% = 20%.
The answer is 31,05 g ethanol.
Unless I've completely forgotten my chemistry, mole fraction is the fraction of moles of a particular solute of the entire number of moles in the solution. So for this question: formula weight of ethanol = 46.07 g/mol formula weight of water = 18.015 g/mol moles of ethanol = 47.5g / 46.07 g/mol = 1.0310 mol moles of water = 850g / 18.015 g/mol = 47.182 mol total number of moles in the solution = 1.0310 + 47.182 = 48.213 mol therefore, the mole fraction of ethanol is 1.0310 mol / 48.213 mol = 0.0214 hope this helps.
The total volume of the solution is 48 mL + 144 mL = 192 mL. The percent by volume of ethanol is calculated as (volume of ethanol / total volume of solution) * 100%. Plugging in the values, we get (48 mL / 192 mL) * 100% = 25%. So, the solution contains 25% ethanol by volume.
1. Extract 959,6 mL from the 99 % solution. 2. Add 40,4 mL water.
Well, darling, you can't make 90% ethanol from 100% ethanol because, sweetie, 100% means it's already pure ethanol. You can dilute it with a calculated amount of water to get to 90%, but you ain't changing that 100% ethanol into something else. So, mix it up with water like a fancy cocktail and voilà, you've got yourself some 90% ethanol.
Volume percent (v/v %) is defined as: volume percent = [(volume of solute) / (volume of solution)] x 100% Volume percent is handy when preparing solutions of liquids. Concentration of a solution can be stated in volume percentages. Be aware that volume of solution is in formula denominator, not volume of solvent. Thus to get 10% v/v solution of ethanol in water you can take 10 ml of ethanol and add enough water to have total 100 ml of resulting solution. It is worth to mention volumes of solute and solvent cannot be simply added to get volume of solution. For instance if you add 10 ml of ethanol to 90 ml of water the volume of the solution will be less than 100 ml.
The mixture water-ethanol is homogeneous.
To increase the concentration of ethanol from 80% to 100%, you can use a process called distillation. By distilling the 80% ethanol solution, you can separate the ethanol from the water and other impurities, resulting in a higher concentration of ethanol.
To prepare 95% ethanol from absolute ethanol, you would need to dilute the absolute ethanol with a calculated amount of distilled water. Since absolute ethanol is 100% pure, you can use the formula C1V1 = C2V2, where C1 is the initial concentration (100%), V1 is the volume of absolute ethanol, C2 is the desired concentration (95%), and V2 is the final volume of the diluted solution. By rearranging the formula and solving for V1, you can determine the volume of absolute ethanol needed to achieve a 95% ethanol solution when mixed with water.