# Relative decrease in vapour pressure of an aqueous NaCl is 0.167. Number of moles of NaCl present in 180 g of H20 is? 1) 2 Mol 2) 1mol 3) 3 Mol 4) 4 mol

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Feb 11, 2018

Here's a way to do it without knowing the multiple choice answers.

I got $\text{2.003 mols NaCl}$. You have given two sig figs, so we can say $\text{2.0 mols NaCl}$.

The decrease in vapor pressure of the solvent $A$ is given by:

$\Delta {P}_{A} = {P}_{A} - {P}_{A}^{\text{*}}$

Assuming the solution follows Raoult's law (i.e. that it is ideal), we get:

$\Delta {P}_{A} = {\chi}_{A \left(l\right)} {P}_{A}^{\text{*" - P_A^"*}}$

$= \left({\chi}_{A \left(l\right)} - 1\right) {P}_{A}^{\text{*}}$

$= - {\chi}_{B \left(l\right)} {P}_{A}^{\text{*}}$

where ${\chi}_{B \left(l\right)}$ is the mol fraction of the solute in the solution phase, ${P}_{A}$ is the vapor pressure of the solvent in the context of the solution, and $\text{*}$ indicates pure solvent.

In this form, it is clear that the change is always negative. If the relative decrease is $0.167$, then we say that

$| \Delta {P}_{A} \frac{|}{P} _ {A}^{\text{*" = 1 - P_A/P_A^"*}} = {\chi}_{B \left(l\right)} = 0.167$

and we see how this turns out to be a fraction; as we have just shown, ${P}_{A} < {P}_{A}^{\text{*}}$ necessarily, whenever any nonvolatile solute is added to the solvent, so $0 < \frac{| \Delta {P}_{A} |}{{P}_{A}^{\text{*}}} < 1$.

By definition,

chi_("NaCl"(l)) = (n_("NaCl"(l)))/(n_("H"_2"O"(l)) + n_("NaCl"(l)))

This means that the mols of $\text{NaCl}$ in the solution phase is found from:

n_("NaCl"(l)) = 0.167(n_("H"_2"O"(l)) + n_("NaCl"(l)))

The mols of water are:

n_("H"_2"O"(l)) = "180 g H"_2"O" xx ("1 mol")/("18.015 g H"_2"O") = "9.992 mols"

Therefore:

$\text{9.992 mols" + n_("NaCl"(l)) = n_"tot}$

and:

n_("NaCl"(l)) = 0.167("9.992 mols" + n_("NaCl"(l)))

Solving for the mols of $\text{NaCl}$ in the solution phase, we get:

$0.167 \left(\text{9.992 mols") = (1 - 0.167)n_("NaCl} \left(l\right)\right)$

color(blue)(n_("NaCl"(l))) = (0.167/(1-0.167))("9.992 mols")

$=$ $\textcolor{b l u e}{\text{2.003 mols}}$

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
SCooke Share
Feb 11, 2018

Use Raoult's Law.
2) 2 moles

#### Explanation:

X_"solvent" = n_"water"/(n_"NaCl" + n_"water")

DeltaP = (X_"solvent" − 1) xx P_"solvent"^0

$\frac{180 g {H}_{2} O}{\frac{18 g}{\text{mol}}} = 10 m o l {H}_{2} O$

X_"solvent" = 10/(n_"NaCl" + 10)

For 2 Moles $N a C l$:
${X}_{\text{solvent}} = \frac{10}{2 + 10} = 0.833$

DeltaP = (X_"solvent" − 1) xx P_"solvent"^0

Relative change $= 0.167 = 1 - 0.833$

• 8 minutes ago
• 13 minutes ago
• 15 minutes ago
• 24 minutes ago
• 12 seconds ago
• 33 seconds ago
• A minute ago
• A minute ago
• 4 minutes ago
• 6 minutes ago
• 8 minutes ago
• 13 minutes ago
• 15 minutes ago
• 24 minutes ago