Question

1f. At an unknown temperature a solution made of (7.740x10^0) g of a non-volatile solute dissolved in 100.0 g of water has a vapor pressure of (5.51x10^1) mm Hg. What is the vapor pressure of pure water (in mm Hg) at this unknown temperature? The molar mass of the solute is (5.360x10^1) g/mol.

1g. A solution is made of two volatile solutes: Chemical A (with a pure vapor pressure of 80.0 mm Hg) and Chemical B (with a pure vapor pressure of 100.0 mm Hg). The solution has a total vapor pressure of (9.88x10^1) mm Hg. If the solution is known to be made from 3.00 mol of A, how many moles of B must there be in the solution?

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Answer 1

1f. Non-volatile solute:

If a non-volatile solute is added to a pure solvent (like water), the resulting pressure of solution (P) is given as:

$P = p^o_{solvent}\chi_{solvent}+p^o_{solute}\chi_{solute}\;\;..(1)$

where is vapor pressure of pure solute

Ps solu

is vapor pressure of pure solvent

Psolvent

are mole fractions of solvent and solute, respectively,

X solvent and xsolute

But, since the solute is non volatile,

therefore

Psolute 0

And, thus,

$P = p^o_{solvent}\chi_{solvent}\;\;..(2)$

Calculation of :

Xsolvent

We know, that:

$\chi_{solvent} = \frac{n_{solvent}}{n_{solvent}+n_{solute}}\;\;..(3)$

$\left ( And,\; n=\frac{Mass(w)}{MolarMass(M)} \right )$

$\chi_{solvent} = \frac{\frac{w_{solvent}}{M_{solvent}}}{\frac{w_{solvent}}{M_{solvent}}+\frac{w_{solute}}{M_{solute}}}$

$\chi_{solvent} = \frac{\frac{100.0g}{18.0g/mol}}{\frac{100.0g}{18.0g/mol}+\frac{7.740g}{53.60g/mol}}$

$\chi_{solvent} = \frac{5.555}{5.555+0.144}$

$\chi_{solvent} = 0.975$

Put this in eq. (2) to solve for

Psolvent

$P = p^o_{solvent}\chi_{solvent}\;\;..(2)$

$p^o_{solvent}=\frac{P}{\chi_{solvent}}$

$p^o_{solvent}=\frac{55.1mmHg}{0.975}$

$p^o_{solvent}=56.5mmHg$

Hence, the vapor pressure of pure water (in mm Hg) at this unknown temperature is 56.5 mmHg

1g. Two volatile solutes:

For a solution made by mixing two volatile solutes, the resulting pressure (P) is:

$P = p^o_{A}\chi_{A}+p^o_{B}\chi_{B}\;\;..(4)$

where $p^o_{A}$ is vapor pressure of pure chemical A

$p^o_{B}$ is vapor pressure of pure chemical B

$\chi_{A} \;and \;\chi_{B}$ are mole fractions of chemical A and B, respectively,

We know that,

$\chi_{A}+\chi_{B}=1\;\;..(5)$

Therefore,

$\chi_{A}=1-\chi_{B}$

With this, (4) becomes,

$P = p^o_{A}\(1-\chi_{B})+p^o_{B}\chi_{B}\;\;..(6)$

$P = p^o_{A}-p^o_A\chi_{B}+p^o_{B}\chi_{B}$

$P = p^o_{A}+(p^o_B-p^o_{A})\chi_{B}$

$P - p^o_{A}=(p^o_B-p^o_{A})\chi_{B}$

$\frac{P - p^o_{A}}{(p^o_B-p^o_{A})}=\chi_{B}$

From (3),

$\frac{98.8mmHg - 80.0mmHg}{100.0mmHg-80.0mmHg}=\frac{n_B}{n_B+3}$

$\frac{18.8mmHg}{20.0mmHg}=\frac{n_B}{n_B+3}$

$\frac{18.8}{20.0}(n_B+3)=n_B$

$0.94(n_B+3)=n_B$

$0.94n_B+2.82=n_B$

$2.82=n_B-0.94n_B$

$2.82=0.06n_B$

$n_B=47mol$

Hence, there are 47 moles of B must there be in the solution