If #"14 g"# of calcium nitrate is dissolved in #"200 g"# of water, and it experiences a #70%# dissociation, what is the solution's vapor pressure if the solvent began at the normal boiling point of water? Hint: the normal boiling point occurs at 760 torr.

Answer 1
#"748 torr"#.
(what is the answer in #"mm Hg"#?)

Vapor pressure of the solution relative to the solvent is given by Raoult's law for ideal solutions:

#P_A = chi_(A(l)) P_A^"*"#

where:

When nonvolatile solute is added to a solvent, it decreases its vapor pressure. Thus, the answer will be less than #"760 torr"#.
In this case, we have a strong electrolyte. Normally, we assume 100% dissociation of strong electrolytes, but in this case there is ion pairing to form #"CaNO"_3^(+)#.
We ignore the #"CaNO"_3^(+)# species in solution for simplicity (which is supported by the fact the solution is dilute so that the ions have a hard time finding each other)
The salt dissociates into #"Ca"^(2+)# and #"NO"_3^(-)#, so:
#"Ca"("NO"_3)_2(aq) -> "Ca"^(2+)(aq) + 2"NO"_3^(-)(aq)#
To calculate the new vapor pressure, we'll need the mol fraction #chi# of the ions or of the water:
#chi_("ions") = n_"ions"/(n_"ions" + n_(H_2O))#
#chi_"ions" + chi_(H_2O) = 1#
We would have three mols of ions, #n_"ions"#, for every one mol of calcium nitrate, #n_(Ca(NO_3)_2)#.
For #"14 g"# of the salt...
#n_(Ca(NO_3)_2) = 14 cancel("g Ca"("NO"_3)_2) xx ("1 mol Ca"("NO"_3)_2)/(164.088 cancel("g Ca"("NO"_3)_2))#
#=# #"0.08532 mols Ca"("NO"_3)_2#

Again, three times the mols of ions result from 100% dissociation, so:

#n_"ions" = 3n_(Ca(NO_3)_2) = "0.2560 mols ions"#

With 70% dissociation, we simply have 70% of the mols, or

#0.7n_"ions" = n_"ions"' = "0.1792 mols"#.
Given #"200 g"# of water, we have:
#200 cancel("g H"_2"O") xx ("1 mol H"_2"O")/(18.015 cancel("g H"_2"O")) = "11.10 mols H"_2"O"#

This gives us a mol fraction of:

#chi_("ions")' = (n_"ions"')/(n_"ions"' + n_(H_2O))#
#= "0.1792 mols"/("0.1792 mols" + "11.10 mols")#
#= 0.01588#
or since all mol fractions for a solution when added together equal #1#,
#chi_(H_2O) = 1 - chi_("ions")' = 0.9841#

Therefore, the solution vapor pressure is:

#color(blue)(P_A) = 0.9841 cdot "760 torr"#
#=# #color(blue)("748 torr")#
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Answer 2

To solve this problem, first, calculate the number of moles of calcium nitrate. Then determine the number of moles of ions produced due to dissociation. Next, find the total number of moles of solute particles. Use the formula for Raoult's law to calculate the vapor pressure of the solution. Finally, apply the given percentage of dissociation to adjust the calculation accordingly. Using Raoult's law:

P_solution = X_solvent * P°_solvent

Where: P_solution = vapor pressure of the solution X_solvent = mole fraction of the solvent P°_solvent = vapor pressure of the pure solvent

Now, calculate the mole fraction of the solvent:

X_solvent = moles of solvent / total moles of solute and solvent

Substitute the values into the equation and solve for the vapor pressure of the solution. Adjust the calculation for the percentage of dissociation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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