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- 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.

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- anonymous

- schrodinger

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- 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

- anonymous

Thank you

- anonymous

@Vocaloid can you help me with another

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- Vocaloid

sure, go for it

- 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

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

- 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

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)

- anonymous

would it be the last option

- anonymous

nvm

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