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1 month ago

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The vapor pressure of benzene, c6h6, is 100.0 torr at 26.1 °c. Assuming raoult's law is obeyed, how many moles of a nonvolatile

solute must be added to 100.0 ml of benzene to decrease its vapor pressure by 10.0% at 26.1 °c? The density of benzene is 0.8765 g>cm3.

1 answer:

Anarel [2.7K]1 month ago

3 0

By Raoult's law:

$p_{solution} = X_{solvent}\times P_{solvent}$ -(1)

Where $p_{solution}$ represents the vapor pressure measured for the solution, $X_{solvent}$ denotes the mole fraction of the solvent, and $P_{solvent}$ specifies the vapor pressure of the pure solvent.

The vapor pressure for benzene is recorded as $100 torr at 26.1^{o} C$ (as given).

Adding a non-volatile solute decreases the vapor pressure by 10.0% (as given).

The vapor pressure of the solution component is thus given by $100 - 10 = 90 torr$ .

Substituting the known values into formula (1):

$90 = mole fraction of benzene \times 100$

$mole fraction of benzene = \frac{90}{100} = 0.9$

The number of moles of benzene is $Volume\times \frac{density}{Molar Mass}$ .

Calculating to find moles of benzene yields $100 cm^{3}\times \frac{0.8765 g/cm^{3}}{78 g/mol} = 1.124 mol$ .

Since, $mole fraction of benzene =\frac{mole of benzene }{mole of benzene +mole of non volatile solute }$

Thus, $0.9 =\frac{1.124 }{1.124+mole of non volatile solute }$

$\frac{1.124}{0.9} =1.124+mole of non volatile solute$

$mole of non volatile solute = 1.249 - 1.124 = 0.125$

Therefore, the moles of non-volatile solute required is $0.125$ .

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An oxygen atom has a mass of and a glass of water has a mass of . Use this information to answer the question below. Be sure you

VMariaS [2867]

Answer:

Number of moles of oxygen atoms that weigh the same as a glass of water = 3.12 moles

Note: This question lacks certain figures. Below is a complete similar question.

An oxygen atom weighs 2.66*10^-23 g and a glass of water weighs 0.050 kg. What is the weight of one mole of oxygen atoms? Round your result to three significant figures. How many moles of oxygen atoms have a weight equal to the weight of a glass of water? Round your answer to two significant figures.

Explanation:

One mole of a substance comprises the Avogadro number of particles, which is 6.02 * 10²³.

Hence, one mole of oxygen atoms contains 6.02*10²³ atoms.

Weight of a single oxygen atom = 2.66*10⁻²³ g

Weight of one mole of oxygen atoms = weight of a single atom multiplied by the number of atoms in one mole.

Weight of one mole of oxygen atoms = 2.66*10⁻²³ g * 6.02*10²³ = 16.01 g

A glass of water weighs = 0.050 kg or 50 g.

To calculate how many moles of oxygen atoms weigh the same as a glass of water (i.e., 50 g), the following formula is applied;

number of moles = mass/molar mass

mass of oxygen atoms = 50 g, molar mass or weight of one mole of oxygen atoms = 16.01 g

Thus, the number of moles of oxygen atoms = 50 g / 16.01 g = 3.12 moles

4 0

1 month ago

You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your di

lorasvet [2668]

Answer:

1) This dilution plan will yield a 200μM solution.

2) This dilution plan will not yield a 200μM solution.

3) This dilution plan will not yield a 200μM solution.

4) This dilution plan will yield a 200μM solution.

5) This dilution plan will yield a 200μM solution.

Explanation:

Convert the initial molarity into molar form as shown.

$500mM = 500mM \times (\frac{1M}{1000M})= 0.5M$

Let's examine the following serial dilution processes.

1)

Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.

Concentration of 500 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{500 mL}= 5 \times 10^{-3}M$

10 mL of this solution is further diluted to 250 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(5 \times 10^{-3}M)(10.0mL)}{250 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

2)

Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.

Concentration of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{100 mL}= 2.5 \times 10^{-2}M$

10 mL of this solution is further diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(2.5 \times 10^{-2}M)(10.0mL)}{1000 mL}= 2.5 \times 10^{-4}M$

Convert μM:

$2.5 \times 10^{-4}M = (2.5 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 250 \mu M$

Thus, this dilution scheme will not yield a 200μM solution.

3)

Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.

Concentration of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{100 mL}= 0.05M$

5 mL of this solution is diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.05M)(5mL)}{1000 mL}= 0.25 \times 10^{-4}M$

Convert μM:

$0.25 \times 10^{-4}M = (0.25 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 25 \mu M$

Thus, this dilution scheme will not yield a 200μM solution.

4)

Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.

Concentration of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5mL)}{250 mL}= 0.01M$

10 mL of this solution is further diluted to 500 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.01M)(10mL)}{500 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

5)

Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.

Concentration of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{250 mL}= 0.02M$

10 mL of this solution is further diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.02M)(10mL)}{1000 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

7 0

1 month ago

An unknown liquid has a mass of 4.25 × 108 mg and a volume of 0.250 m3. what is the density of the liquid in units of g/ml?

VMariaS [2867]

Density is calculated as mass divided by volume.
Step one:
Convert m³ to ml.
1 m³ = 1,000,000 ml
0.250 m³ x 1,000,000 = 250,000 ml
Step two: Convert mg to g.
1 mg = 0.001 g, hence 4.25 x 10^8 mg equals 0.459 g.
Consequently, the density comes out to be 0.459 g/250,000 = 1.836 x 10^-6 g/ml.

8 0

1 month ago

Read 2 more answers

What is the stoichiometric ratio between BaCl2 and NaCl

eduard [2652]

<span>BaCl2 + Na2SO4 --> BaSO4 + 2NaCl In this reaction, 1.0 g of BaCl2 and 1.0 g of Na2SO4 are present. We need to identify the limiting reactant. "First, convert grams to moles" 1.0 g BaCl2 * (1 mol BaCl2 / 208.2 g BaCl2) = 4.8 x 10^-3 mol BaCl2 1.0 g Na2SO4 * (1 mol Na2SO4 / 142.04 g Na2SO4) = 7.0 x 10^-3 mol Na2SO4 (7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2) = 1.5 mol Na2SO4 per mol BaCl2 "Using this ratio to compare with the balanced equation, BaCl2 + Na2SO4 --> BaSO4 + 2NaCl" The balanced equation indicates that 1 mol of BaCl2 reacts with 1 mol of Na2SO4. However, we found that 1.5 mol of Na2SO4 is available for each mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>

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1 month ago

"Enter your answer in the provided box. A person's blood alcohol (C2H5OH) level can be determined by titrating a sample of blood

Tems11 [2638]

Answer:

0.1714 (w/w) %

Explanation:

Utilizando la ecuación:

16H+(aq) + 2Cr2O72−(aq) + C2H5OH(aq) → 4Cr3+(aq) + 2CO2(g) + 11H2O(l)

Se emplean 2 moles de ion dicromato (Cr₂O₇²⁻) para titular 1 mol de alcohol (C₂H₅OH)

35.46mL = 0.03546L de una solución de Cr₂O₇²⁻ a 0.05961M utilizada para alcanzar el punto de equivalencia en la titulación contiene:

0.03546L ₓ (0.05961 moles Cr₂O₇²⁻ / L) = 2.114x10⁻³ moles Cr₂O₇²⁻

Dado que 2 moles de dicromato reaccionan por cada mol de alcohol, los moles de alcohol en la muestra de plasma son:

2.114x10⁻³ moles Cr₂O₇²⁻ ₓ ( 1 mol C₂H₅OH / 2 moles Cr₂O₇²⁻) = 1.0569x10⁻³ moles de C₂H₅OH

Como la masa molar del alcohol es 46.07g/mol, la masa de alcohol es:

1.0569x10⁻³ moles de C₂H₅OH ₓ (46.07g / mol) = 0.04869g de C₂H₅OH

Por lo tanto, el porcentaje en masa de alcohol en sangre utilizando los 28.40g de plasma es:

(0.04869g de C₂H₅OH / 28.40g) × 100 = 0.1714 (w/w) %

7 0

1 month ago