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What is the vapor pressure of the solution if 35.0 g of water is dissolved in 100.0 g of ethyl alcohol at 25 ∘C? The vapor press

masya89 [10]

Answer: The vapor pressure of the solution is 43.55 mmHg

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For water:

Given mass of water = 35.0 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{35.0g}{18g/mol}=1.944mol$

For ethyl alcohol:

Given mass of ethyl alcohol = 100.0 g

Molar mass of ethyl alcohol = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of ethyl alcohol}=\frac{100.0g}{46g/mol}=2.174mol$

Total moles of solution = [1.944 = 2.174] moles = 4.118 moles

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

For water:

$\chi_{\text{water}}=\frac{n_{\text{water}}}{n_{\text{water}}+n_{\text{ethyl alcohol}}}$

$\chi_{water}=\frac{1.944}{4.118}=0.472$

For ethyl alcohol:

$\chi_{\text{ethyl alcohol}}=\frac{n_{\text{ethyl alcohol}}}{n_{\text{water}}+n_{\text{ethyl alcohol}}}$

$\chi_{\text{ethyl alcohol}}=\frac{2.174}{4.118}=0.528$

Dalton's law of partial pressure states that the total pressure of the system is equal to the sum of partial pressure of each component present in it.

To calculate the vapor pressure of the solution, we use the law given by Dalton, which is:

$P_T=\sum_{i=1}^n (p_i\times \chi_i)$

Or,

$P_T=[(p_{\text{water}}\times \chi_{\text{water}})+(p_{\text{ethyl alcohol}}\times \chi_{\text{ethyl alcohol}}$

We are given:

Vapor pressure of water = 23.8 mmHg

Vapor pressure of ethyl alcohol = 61.2 mmHg

Putting values in above equation, we get:

$p_T=[(23.8\times 0.472)+(61.2\times 0.528)]\\\\p_T=43.55mmHg$

Hence, the vapor pressure of the solution is 43.55 mmHg

4 0

4 years ago

See attached image down below

cestrela7 [59]

I think the answer will be 60 and the reason why is because

6 0

3 years ago

Copper metal (Cu) reacts with silver nitrate (AgNO3) in aqueous solution to form Ag and Cu(NO3)2. An excess of AgNO3 is present.

leonid [27]

Stoichiometry time! Remember to look at the equation for your molar ratios in other problems.

31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag         6851.65
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻   →   ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ = 107.9 g Ag
       ∅        | 63.5 g Cu | 1 mol Cu | 1 mol Ag              63.5

There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.

3 0

3 years ago

Read 2 more answers

A 2.1−mL volume of seawater contains about 4.0 × 10−10 g of gold. The total volume of ocean water is about 1.5 × 1021 L. Calcula

Fiesta28 [93]

Answer:

Total worth of gold in the ocean = $5,840,000,000,000,000

Explanation:

As stated in the question above, 4.0 x 10^-10 g of gold was present in 2.1mL of ocean water.

Therefore, In 1 L of ocean water there will be,

(4.0 x 10^-10)/0.0021

= 1.9045 x 10^-7 g of gold per Liter of ocean water.