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1. convert 185g of water to kg 2. divide moles of NaCl by kg of water to get molality of NaCl 3. use the formula boiling point elevation = (n)(Kb)(m) where n is the number of particles, Kb is 0.51, and m is molality of NaCl 4. add boiling point elevation to 100degrees Celcius to get the final boiling point
sure, go for it
A solution is made by dissolving 21.5 grams of glucose (C6H12O6) in 255 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.
well, luckily for you, boiling point elevation and freezing point depression are almost the same in terms of the process 1. convert 255 g of water to kg 2. convert 21.5g of glucose to moles by dividing by the molar mass of glucose 3. find the molality of glucose by dividing moles of glucose/kg of water 4. use the formula freezing point depression = (n)(Kf)(m) where n is the number of particles (n=1 in this case, since glucose doesn't dissociate), Kf is -1.86, and m is molality of glucose (obtained in step 3) 4. subtract the result from 0 to find the final freezing point
Thank you. Can you check my answer on this problem Which of the following aqueous solutions will have the lowest vapor pressure at 25°C? 1.5 M C6H12O6 1.5 M LiNO3 1.0 M Al2O3 1.0 M CaF2 I chose the first one
hm, not quite lowest vapor pressure should be the solution with the highest number of particles for each answer choice, multiply (number of particles)*(concentration)
would it be the last option