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1 year ago

A solution is prepared by mixing 200.0 g of water, H2O, and 300.0 g of

Chemistry

1 answer:

VikaD [51]1 year ago

7 0

Answer:

Mole Fraction (H₂O) = 0.6303

Mole Fraction (C₂H₅OH) = 0.3697

Explanation:

(Step 1)

Calculate the mole value of each substance using their molar masses.

Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol

Molar Mass (H₂O): 18.014 g/mol

200.0 g H₂O 1 mole
--------------------- x ------------------ = 11.10 moles H₂O
18.014 g

Molar Mass (C₂H₅OH): 2(12.011 g/mol) + 6(1.008 g/mol) + 15.998 g/mol

Molar Mass (C₂H₅OH): 46.068 g/mol

300.0 g C₂H₅OH 1 mole
---------------------------- x -------------------- = 6.512 moles C₂H₅OH
46.068 g

(Step 2)

Using the mole fraction ratio, calculate the mole fraction of each substance.

moles solute
Mole Fraction = ------------------------------------------------
moles solute + moles solvent

11.10 moles H₂O
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH

Mole Fraction (H₂O) = 0.6303

6.512 moles C₂H₅OH
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH

Mole Fraction (C₂H₅OH) = 0.3697

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3 years ago

What is the molality, m, of an aqueous solution of ammonia that is 12.83 M NH3 (17.03 g/mol)? This solution has a density of 0.9

daser333 [38]

Answer:

Molality = 18.5 m

Explanation:

Let's analyse data. We want to determine molality which means mol of solute / 1kg of solvent. (Hence we need, the moles of solute and the mass of solvent in kg)

12.83 M means molarity → mol of solute in 1L of solution

Density refers always to solution → Mass of solution / Volume of solution

1L = 1000 mL

We can determine the mass of solution with density

0.9102 g/mL = Mass of solution / 1000 mL

Mass of solution = 0.9102 g/mL . 1000 mL → 910.2 g

Let's convert the moles of solute (NH₃) to mass

12.83 mol . 17.03 g/ 1 mol = 218.5 g

We can apply this knowledge:

Mass of solution = Mass of solvent + Mass of solute

910.2 g = Mass of solvent + 218.5 g

910.2 g - 218.5 g = 691.7 g → Mass of solvent.

Let's convert the mass in g to kg

691.7 g . 1kg / 1000 g = 0.6917kg

We can determine molalilty now → 12.83 mol / 0.6917kg

Molality = 18.5 m

6 0

2 years ago

How the change in pressure affect the volume of the gas

Strike441 [17]

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3 years ago

What is the change in internal energy ( ΔU ) of the system if q = –8 kJ and w = –1 kJ for a certain process?

Radda [10]

Answer:

Change in internal energy (ΔU) = -9 KJ

Explanation:

Given:

q = –8 kJ [Heat removed]

w = –1 kJ [Work done]

Find:

Change in internal energy (ΔU)

Computation:

Change in internal energy (ΔU) = q + w

Change in internal energy (ΔU) = -8 KJ + (-1 KJ)

Change in internal energy (ΔU) = -8 KJ - 1 KJ

Change in internal energy (ΔU) = -9 KJ

6 0

3 years ago

How many milliliters of an aqueous solution of 0.170 M ammonium carbonate is needed to obtain 16.1 grams of the salt

Citrus2011 [14]

There will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.There will be needed mL of

Why?

In order to calculate how many milliliters are needed to obtain 16.1 grams of the salt given its concentration, we first need to find its chemical formula which is the following:

$(NH_{4})2CO_{3}$

Now that we know the chemical formula of the substance, we need to find its molecular mass. We can do it by the following way:

$N_{2}=14g*2=28g\\\\2H_{4}=2*1g*4=8g\\\\C=12.01g*1=12.01g\\\\O_{3}=15.99g*3=47.97g$

We have that the molecular mass of the substance will be:

$MolecularMass=\frac{28g+8g+12.01g+47.97g}{mol}=95.98\frac{g}{mol}$

Therefore, knowing the molecular mass of the substance, we need to calculate how many mols represents 16.1 grams of the same substance, we can do it by the following way:

$mol_{(NH_{4})2CO_{3}=\frac{mass_{(NH_{4})2CO_{3}}}{molarmass_{(NH_{4})2CO_{3}}}$

$mol_{(NH_{4})2CO_{3}=\frac{16.1g}{95.98\frac{g}{mol}}=0.167mol$

Finally, if we need to calculate how many milliliters are needed, we need to use the following formula:

$M=\frac{moles_{solute}}{volume_{solution}}$

$M=\frac{moles_{solute}}{volume_{solution}}\\\\volume_{solution}=\frac{moles_{solute}}{M}$

Now, substituting and calculating, we have:

$volume_{solution}=\frac{0.167mol}{0.170\frac{mol}{L}}\\\\volume_{solution}=0.982L=0.982L*1000=982.35mL$

Henc, there will be needed 982.35 mL of solution to obtain 16.1 grams of the salt.

Have a nice day!

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3 years ago