I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
Report (any type)/Brief report
22/09/2021
Here’s what I got.
In order to be able to calculate the , mole fraction, and of the solution, you first need to pick a volume sample.
Since is defined as moles of per liters of solution, a ##”1.00-L”## sample will make the calculations easier.
So, let’s say that we have a ##”1.00-L”## sample of this solution. You can use its given to determine the mass of this sample
##1.00 color(red)(cancel(color(black)(“L”))) * (1000 color(red)(cancel(color(black)(“mL”))))/(1color(red)(cancel(color(black)(“L”)))) * “1.036 g”/(1color(red)(cancel(color(black)(“mL”)))) = “1036 g”##
Now, you know that this solution is ##12.00%## ammonium chloride by mass. This means that every ##”100 g”## of solution will contain ##”12.0 g”## of ammonium chloride, ##”NH”_4″Cl”##.
In your case, the sample will contain
##1036 color(red)(cancel(color(black)(“g solution”))) * (“12.00 g NH”_4″Cl”)/(100color(red)(cancel(color(black)(“g solution”)))) = “124.32 g NH”_4″Cl”##
Next, use ammonium chloride’s molar mass to determine how many moles of the compound will be present in this many grams
##124.32 color(red)(cancel(color(black)(“g”))) * (“1 mole NH”_4″Cl”)/(53.49color(red)(cancel(color(black)(“g”)))) = “2.3242 moles NH”_4″Cl”##
This means that the molarity of the solution will be
##color(blue)(c = n/V)##
##c = “2.3242 moles”https://studydaddy.com/”1.00 L” = color(green)(“2.324 M”)##
To get the mole fraction of ammonium chloride, you need to know the total number of moles present in the solution sample. More specifically, you need to figure out how many moles of water, the , are present in this sample.
To do that, use the mass of the solution and the mass of ammonium chloride
##m_”water” = m_”solution” – m_”ammonium chloride”##
##m_”water” = “1036 g” – “124.32 g” = “911.68 g H”_2″O”##
Use water’s molar mass to figure out how many moles can be found in this many grams
##911.68 color(red)(cancel(color(black)(“g”))) * (“1 mole H”_2″O”)/(18.015 color(red)(cancel(color(black)(“g”)))) = “50.607 moles H”_2″O”##
The total number of moles present in solution will thus be
##n_”total” = 2.3242 + 50.607 = “52.931 moles”##
fraction of ammonium chloride, which is equal to the number of moles of ammonium chloride divided by the total number of moles in solution, will be
##chi_(NH_4Cl) = (2.3242 color(red)(cancel(color(black)(“moles”))))/(52.931color(red)(cancel(color(black)(“moles”)))) = color(green)(0.04391)##
Finally, is defined as moles of solute divided by kilograms of solvent. Since this sample contain ##”911.68 g”## of water, its molality will be
##color(blue)(b = n_”solute”/m_”solvent”)##
##b = “2.3242 moles”/(911.68 * 10^(-3)”kg”) = color(green)(“2.549 molal”)##
The answers are rounded to four , the number of sig figs you have for the and of the solution.
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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