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Mathematics 13 Online
OpenStudy (anonymous):

I KNOW THIS A DIFFERENT SUBJECT **Please just don't make me go through a long process. I just need to finish the class.** A solution is made by dissolving 3.8 moles of sodium chloride (NaCl) in 185 grams of water. If the molal boiling point constant for water (Kb) is 0.51 °C/m, what would be the boiling point of this solution? Show all of the work needed to solve this problem.

Vocaloid (vocaloid):

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

OpenStudy (anonymous):

Thank you

OpenStudy (anonymous):

@Vocaloid can you help me with another

Vocaloid (vocaloid):

sure, go for it

OpenStudy (anonymous):

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.

Vocaloid (vocaloid):

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

OpenStudy (anonymous):

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

Vocaloid (vocaloid):

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)

OpenStudy (anonymous):

would it be the last option

OpenStudy (anonymous):

nvm

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