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3 years ago

A solution is made by dissolving

Chemistry

1 answer:

barxatty [35]3 years ago

8 0

Answer:

4.52 mol/kg

Explanation:

Given data:

Mass of lithium fluoride = 22.1 g

Mass of water = 188 g

Molality = ?

Solution:

Molality:

It is the number of moles of solute into kilogram of solvent.

Formula:

Molality = number of moles of solute / kilogram solvent

Mathematical expression:

m = n/kg

Now we will convert the grams of LiF into moles.

Number of moles = mass/ molar mass

Number of moles = 22.1 g/ 26 g/mol

Number of moles = 0.85 mol

Now we will convert the g of water into kg.

Mass of water = 188 g× 1kg/1000 g = 0.188 kg

Now we will put the values in formula.

m = 0.85 mol / 0.188 kg

m = 4.52 mol/kg

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A 0.4021-g sample of a purified organic acid was dissolved in water and titrated potentiometrically. A plot of the data revealed

Archy [21]

Answer:

The molar mass of the acid is 167.5 g/mol

Explanation:

A 0.4021-g sample of a purified organic acid was dissolved in water and titrated potentiometrically. A plot of the data revealed a single end point after 19.31 mL of 0.1243 M NaOH had been introduced. Calculate the molecular mass of the acid.

Step 1: Data given

Mass of the sample of a purified organic acid = 0.4021 grams

Molarity = 0.1243 M

Volume needed to reach the end point = 19.1 mL = 0.01931 L

Step 2: Calculate the number of moles NaOH

Moles NaOH = molarity NaOH * volume

Moles NaOH = 0.1243 M * 0.01931 L

Moles NaOH = 0.00240 moles

Step 3: Calculate moles of the acid

We'll need 0.00240 moles of acid to neutralize 0.00240 moles of NaOH ( it's a single end point)

Moles acid = 0.00240 moles

Step 4: Calculate molar mass of the acid

Molar mass = mass / moles

Molar mass = 0.4021 grams / 0.00240 moles

Molar mass = 167.5 g/mol

The molar mass of the acid is 167.5 g/mol

6 0

3 years ago

What mass of carbon dioxide gas occupies a volume of 8.13 L

Natasha_Volkova [10]

Answer:

23.76g

Explanation:

To solve for the mass of carbon dioxide (CO2), let us first obtain the number of mole of CO2.

From the question, the following were obtained:

V = 8.13L

P = 204kPa = 204000Pa

Recall: 101325Pa = 1atm

204000Pa = 204000/101325 = 2atm

T = 95°C = 95 + 273 = 368K

R = 0.082atm.L/K/mol

n =?

PV = nRT

n = PV /RT

n = 2 x 8.13/ 0.082 x 368

n = 0.54mole

Now let us convert 0.54mole of CO2 to gram. This is illustrated below:

Molar Mass of CO2 = 12 + (16x2) = 12 + 32 = 44g/mol

Number of mole of CO2 = 0.54mole

Mass of CO2 =?

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass of CO2 = 0.54 x 44

Mass of CO2 = 23.76g

7 0

4 years ago

78.6 grams of O2 and 67.3 grams of F2 are placed in a container with a volume of 40.6 L. Find the total pressure if the gasses a

saul85 [17]

1) List the known and unknown quantities.

Sample: O2.

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Sample: F2.

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

2) Find the pressure of O2.

2.1- List the known and unknown quantities.

Sample: O2.

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

2.2- Convert grams of O2 to moles of O2.

The molar mass of O2 is 31.9988 g/mol.

$mol\text{ }O_2=78.6\text{ }g*\frac{1\text{ }mol\text{ }O_2}{31.9988\text{ }g\text{ }O_2}=2.46\text{ }mol\text{ }O_2$

2.3- Set the equation.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

$PV=nRT$

2.4- Plug in the known quantities and solve for P.

$(P)(40.6\text{ }L)=(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)$

.

$P_{O_2}=\frac{(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)}{40.6\text{ }L}$ $P_{O_2}=1.57\text{ }atm$

The pressure of O2 is 1.57 atm.

3) Find the pressure of F2.

3.1- List the known and unknown quantities.

Sample: F2.

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

3.2- Convert grams of F2 to moles of F2.

The mmolar mass of F2 is 37.9968 g/mol.

$mol\text{ }F_2=67.3\text{ }g\text{ }F_2*\frac{1\text{ }mol\text{ }F_2}{37.9968\text{ }g\text{ }F_2}=1.77\text{ }mol\text{ }F_2$

3.3- Set the equation.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

$PV=nRT$

3.4- Plug in the known quantities and solve for P.

$(P)(40.6\text{ }L)=(1.77\text{ }mol\text{ }F_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)$

.

$P_{F_2}=\frac{(1.77molF_2)(0.082057L*atm*K^{-1}*mol^{-1})(316.28K)}{40.6\text{ }L}$ $P_{F_2}=1.13\text{ }atm$

The pressure of F2 is 1.13 atm.

4) The total pressure.

Dalton's law - Partial pressure. This law states that the total pressure of a gas is equal to the sum of the individual partial pressures.

4.1- Set the equation.

$P_T=P_A+P_B$

4.2- Plug in the known quantities.

$P_T=1.57\text{ }atm+1.13\text{ }atm$ $P_T=2.7\text{ }atm$

The total pressure in the container is 2.7 atm.

5 0

1 year ago

A chemistry student weighs out of acetic acid into a volumetric flask and dilutes to the mark with distilled water. He plans to

xeze [42]

Answer:

Your question is not complete, but use this answer as a guide for your solution.

Question: A chemistry student weighs out 0.112g of acetic acid (HCH₃CO₂) into a 250. mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1600 M NaOH solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Be sure your answer has the correct number of significant digits.

Answer: Volume of NaOH is 11.6 mL

Explanation:

The reaction of acetic acid with NaOH is as follows:

CH3COOH + NaOH -----> CH3COONa + H2O

M1V1 = M2V2

Here M1 V1 are molarity and volume of acetic acid.

M2, V2 are molarity and volume of NaOH.

Number of moles of acetic acid:

0.112 g CH3COOH × (1 mol / 60.05 g) = 0.001865 mol

Molarity = moles of solute / Liters of solution

Molarity = 0.001865 mol / 0.250 L = 0.00746 M

Hence,

M1 = 0.00746 M

V1 = 250 mL

M2 = 0.160 M

V2 = ?

V2 = M1V1 / M2

V2 = 0.00746 M × 250 mL / 0.160 M

V2 = 11.6 mL

Hence the volume of NaOH is 11.6 mL

4 0

3 years ago

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sineoko [7]

Answer:

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3 years ago