an aqueous solution is prepared by dissolving 2.66g of aluminum chloride in 100g of waterwhat is the freezing point (?T) of this solution?please please answer.and can you explain how to use the equation for freezing point??
i think using spray paint or fabric paint (even if it is diluted) would not be the best choice because it would cause the fibers of the fur to stick togetherThis might give it a matted look from far awayAlso you will never get a pure black when dying it (black are just really dark colors)If you dye it a darker color you run the risk of not getting a black cat but whatever color the black is based on (for example a red based black might dye it a dark red)This is important because fabrics can only be dyed so farsince it is already dyed you probably cannot get it to a black, but just a dark colorDon't bother with Dylon because it is basically high quality RIT dyeIf you do use fabric paint I recommend Jacquard, which you can find in craft stores like Michaels and JoAnne's, it's easier to find than dyna flo
Molecular weight of aluminium chloride (AlCl3)27 + 35.5×3133.5 Mass of AlCl3 present in solution2.66 g/mol Therefore, noof moles of AlCl3 present in solution2.66/133.5 0.0199250936 molesNow, the molality of the solution is equal to the noof moles of solute (AlCl3) present in 1 kg, i.e 1000 g of the solvent Noof moles of aluminium chloride present in 100g of water0.0199250936 Therefore, noof moles of AlCl3 in 1000g of water (0.0199250936×1000)/100 0.199250936 Therefore, molality of the solution (m)0.199250936 The equation used to determine the depression in freezing point is given by ?Tf iKf × m, where ?Tf is the freezing point depression, Kf is the cryoscopic constant for water, and m is the molality of the solution, and i is the Van't Hoff factorAlCl3 dissociates as AlCl3?Al3+ + 3Cl- Assuming complete dissociation, i4 Cryoscopic constant for water (Kf)1.86 °C/m Therefore, depression in freezing point (?Tf) iKf × m4×1.86 °C/m × 0.1992509361.48242696°C freezing point of pure water 0°C Therefore, the freezing point of the aqueous solution 0°C-?Tf0°C-1.48242696°C -1.48242696°C Thus, the solution freezes at -1.48242696°C Edit: Oh gosh, I can't believe that the noof significant digits can be such a great issue that it has to be pointed out! after all, whether they are included or not, it does not matterYeah, I know that it's sufficient to give the answer as upto 2 places of decimal, but I calculated the answer, using the google calculator, and pasted the same without bothering to round it up! Anyway, never mind.
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This Site Might Help YouRE: Is polyester jacquard a soft fabric? I want to buy a duvet cover but it's in polyester jacquardIs it that noisy kind of fabric? Or is it soft like cotton?
i think using spray paint or fabric paint (even if it is diluted) would not be the best choice because it would cause the fibers of the fur to stick togetherThis might give it a matted look from far awayAlso you will never get a pure black when dying it (black are just really dark colors)If you dye it a darker color you run the risk of not getting a black cat but whatever color the black is based on (for example a red based black might dye it a dark red)This is important because fabrics can only be dyed so farsince it is already dyed you probably cannot get it to a black, but just a dark colorDon't bother with Dylon because it is basically high quality RIT dyeIf you do use fabric paint I recommend Jacquard, which you can find in craft stores like Michaels and JoAnne's, it's easier to find than dyna flo
Molecular weight of aluminium chloride (AlCl3)27 + 35.5×3133.5 Mass of AlCl3 present in solution2.66 g/mol Therefore, noof moles of AlCl3 present in solution2.66/133.5 0.0199250936 molesNow, the molality of the solution is equal to the noof moles of solute (AlCl3) present in 1 kg, i.e 1000 g of the solvent Noof moles of aluminium chloride present in 100g of water0.0199250936 Therefore, noof moles of AlCl3 in 1000g of water (0.0199250936×1000)/100 0.199250936 Therefore, molality of the solution (m)0.199250936 The equation used to determine the depression in freezing point is given by ?Tf iKf × m, where ?Tf is the freezing point depression, Kf is the cryoscopic constant for water, and m is the molality of the solution, and i is the Van't Hoff factorAlCl3 dissociates as AlCl3?Al3+ + 3Cl- Assuming complete dissociation, i4 Cryoscopic constant for water (Kf)1.86 °C/m Therefore, depression in freezing point (?Tf) iKf × m4×1.86 °C/m × 0.1992509361.48242696°C freezing point of pure water 0°C Therefore, the freezing point of the aqueous solution 0°C-?Tf0°C-1.48242696°C -1.48242696°C Thus, the solution freezes at -1.48242696°C Edit: Oh gosh, I can't believe that the noof significant digits can be such a great issue that it has to be pointed out! after all, whether they are included or not, it does not matterYeah, I know that it's sufficient to give the answer as upto 2 places of decimal, but I calculated the answer, using the google calculator, and pasted the same without bothering to round it up! Anyway, never mind.
