Gen chem question!

[flin5845]

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When 25.5 grams of a nonvolatile nonelectrolyte are placed into 500 grams of water, the boiling point of the solution at one atmosphere is 101.56 ºC. What is the molecular weight of the solute? (Kb of H2O = 0.52 K·kg/mol)
A.102 g/mol
B.51 g/mol
C.34 g/mol
D.17 g/mol
E.8.5 g/mol

The explanation states that deltaT is apparently 3 times the boiling point elevation constant. So since 25.5g of solute was dissolved in .5kg of solvent, then 51g of solute would be required to make a 3m solution with one kg. Thus 3 moles of solute per kg of solvent is equal to 51g of solute per kg. SO the formula weight is 51/3= 16grams/mol

Can someone please explain this question to me the explanation was confusing and made no sense. I do not understand why the had to make a 1kg solvent and how does 3 moles of solute equal 51g of solute?

 

[RalphMouth]

delta T = 1.56

delta T = i(1 in this case since nonelectrolyte)*K*m

1.56 = 0.53 * m

m = 3

m = moles of solute/ Kg solvent

3= x/0.5

x = 1.5 moles

25.5/MW = 1.5 moles, MW = 17

They just shortened the stoichemistry a bit and explained it a little akwardly.

 

[flin5845]

delta T = 1.56

delta T = i(1 in this case since nonelectrolyte)*K*m

1.56 = 0.53 * m

m = 3

m = moles of solute/ Kg solvent

3= x/0.5

x = 1.5 moles

25.5/MW = 1.5 moles, MW = 17

They just shortened the stoichemistry a bit and explained it a little akwardly.

Thank you! That makes so much more sense then what the explanation was trying to do!

 

[flin5845]

I have one more that I do not like the way the tried to explain it.

What is the Molarity of K2SO4 in an aqueous solution created by adding 112 g of solid KOH to 500mL of .5M H2SO4. (Answer is .5M)

 

[RalphMouth]

112g KOH = 2 moles of KOH. Dissociated that represents 2 moles of K+ and 2 moles of OH-.

0.5 M H2SO4 * 0.5 L = 0.25 moles H2SO4. Looking at the equation, there is only one mole of sulfate ion, therefore .25 moles of SO4-2.

In K2SO4, you can see that SO4-2 will be limiting (just compare the coefficients of K and sulfate vs how much is actually present 2 mol, 0.25 mol).

0.25 moles of K2SO4/ 0.5 L of solution = 0.5 M.

 

[addis4ever]

where do you find these problems?

 

[flin5845]

where do you find these problems?

They are from a Kaplan practice exam. Here is one more I am not sure why my way did not work.

MnO4- + I- + H+→ Mn2+ + H2O + I2

If 150 mg of KI is reacted and 108.5 mg of I2 is produced, what is the percent yield for the reaction?

I balanced the equation and basically have 10 I- → 5 I2.... since we have 150 mg of KI I just subtracted the MW of K from that to get I- which was 110mg and then convert that into mols of I- and use the ratio to find the actual amount of I2. ....I feel so dumb haha

 

[rah08e]

When 25.5 grams of a nonvolatile nonelectrolyte are placed into 500 grams of water, the boiling point of the solution at one atmosphere is 101.56 ºC. What is the molecular weight of the solute? (Kb of H2O = 0.52 K·kg/mol)
A.102 g/mol
B.51 g/mol
C.34 g/mol
D.17 g/mol
E.8.5 g/mol

The explanation states that deltaT is apparently 3 times the boiling point elevation constant. So since 25.5g of solute was dissolved in .5kg of solvent, then 51g of solute would be required to make a 3m solution with one kg. Thus 3 moles of solute per kg of solvent is equal to 51g of solute per kg. SO the formula weight is 51/3= 16grams/mol

Can someone please explain this question to me the explanation was confusing and made no sense. I do not understand why the had to make a 1kg solvent and how does 3 moles of solute equal 51g of solute?

Why did RalphMouth use "0.53" for K? Shouldn't K be "0.52" as it's given in the problem?

 

[flin5845]

Why did RalphMouth use "0.53" for K? Shouldn't K be "0.52" as it's given in the problem?

I think it was just a typo....either way 3 mols is correct.

 

[flin5845]

MnO4- + I- + H+→ Mn2+ + H2O + I2

If 150 mg of KI is reacted and 108.5 mg of I2 is produced, what is the percent yield for the reaction?

Anyone know about this one?

 

[LetsGo2DSchool]

MnO4- + I- + H+→ Mn2+ + H2O + I2

If 150 mg of KI is reacted and 108.5 mg of I2 is produced, what is the percent yield for the reaction?

Anyone know about this one?

First balance the equation.

Then convert 150 mg KI into moles.

That will give you the number of moles of I- reacted.

Then calculate the number of moles of I2 produced based on the balanced equation.

Convert the moles of I2 produced into g. (Actual yield)

Then divide actual by theoretical (108.5g) and that will be percent yield.

 

[flin5845]

First balance the equation.

Then convert 150 mg KI into moles.

That will give you the number of moles of I- reacted.

Then calculate the number of moles of I2 produced based on the balanced equation.

Convert the moles of I2 produced into g. (Actual yield)

Then divide actual by theoretical (108.5g) and that will be percent yield.

Ok so I got a balanced equation of: 10I- ----> 5I2.

I converted 150mg to .15g of KI then divided that by the MW of KI (165g) and got 9x10^-4 mol of I-...

Then since it is a 2:1 ratio (of I- to I2) I divided 9x10^-4 mol of I- by 2 to get the # of mols of I2 which came out to 4.5x10^-4.

Then I multiplied 4.5x10^-4 times the MW of I2 (250g) to get the number of grams of I2. Which was .1125g of I2. This would be my actual amount of I2 in grams. So I multiplied by 1000 to get mg. which was 112.5mg...

but 112.5mg/108.5mg=103% yield!

WHERE AM I GOING WRONG?!?!

 

[RalphMouth]

Your actual yield is 108.5 mg (given), not your theoretical yield.

Your theoretical yield is the calculation of 112.5 mg (I haven't actually done the math but your calculations look correct for that). It is how much you should produce from 150 mg of KI, that's why it's calculated.

So just flip the numerator and denominator.

And yes for problem 1 that was a typo on my part.

 

[flin5845]

Your actual yield is 108.5 mg (given), not your theoretical yield.

Your theoretical yield is the calculation of 112.5 mg (I haven't actually done the math but your calculations look correct for that). It is how much you should produce from 150 mg of KI, that's why it's calculated.

So just flip the numerator and denominator.

And yes for problem 1 that was a typo on my part.

Thanks man I appreciate it!

 

[baywatch123]

i know this post this super old..... but in the answers they don't convert the MM of I and KI. they just have it as

(127 mg I)/(166 mg KI) and times 150 mg
why is this?

nvmm... the ratio's just cancel out