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3 years ago

A solution of a sugar with the chemical formula (c12h22o11) is prepared by dissolving 8.45 g in 250.0 ml of water at 25

2 answers:

6 0

A. Concentration in terms of molarity

First, calculate for the molar mass of the given solution, C12H22O11.
m = 12(12) + 22(1) + 11(16) = 232
Calculate for the number of moles.

n = 8.45 g / 232 g/mol = 0.036422 moles

Divide the determined number of moles by volume of solution in L.
V(solution) = (250 mL)(1 L/1000 mL) = 0.25 L

The concentration in molarity is determined by dividing the number of moles by the volume of solution in liters.
M = 0.03622 moles / 0.25 L = 0.146 moles/L

b. Concentration in terms of molality
Calculate the mass of the solvent in kilogram given the volume and its density.
m(solvent) = (1.21 g/cm3)(250 mL)(1 kg/1000 g) = 0.3025 kg
Divide the calculated number of moles by the mass of the solvent in kilograms.

molality = 0.036422 moles/0.3025 kg = 0.12 molal

c. Concentration in terms of weight percent.
The total weight of the solution is equal to 302.5 g.

weight percent of solute = (8.45 g / 302.5 g)(100%) = 2.79%

d. Weight percent in ppm.
weight percent of solute x (10000 ppm/1%)
= (2.79%)(10000 ppm/1%) = 27933.88 ppm

tigry1 [53]3 years ago

6 0

Answer : The molarity, molality, weight percent and ppm are, 0.0987 mole/L, 0.0839 mole/Kg, 2.79% and 27933.8843 respectively.

Solution : Given,

Mass of solute $(C_{12}H_{22}O_{11})$ = 8.45 g

Molar mass of solute $(C_{12}H_{22}O_{11})$ = 342.3 g/mole

Volume of solvent (water) = volume of solution = 250 ml

Density of solution = $1.21g/cm^3=1.21g/ml$

Calculation for molarity

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}$

$Molarity=\frac{8.45g\times 1000}{342.3g/mole\times 250ml}=0.0987mole/L$

Now we have to calculate the mass of solution.

$\text{Mass of solution}=Density\times Volume=1.21g/ml\times 250ml=302.5g$

Now we have to calculate the mass of solvent.

$\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}=302.5-8.45=294.05g$

Calculation for molality

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent in g}}$

$Molality=\frac{8.45g}\times 1000}{342.3g/mole\times 249.05g}=0.0839mole/Kg$

Calculation for weight percent

$weight\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{8.45g}{302.5g}\times 100=2.79\%$

Calculation for ppm

$ppm=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6=\frac{8.45g}{302.5g}\times 10^6=27933.8843$

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The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

3 years ago

A student is adding DI water to a volumetric flask to make a 50% solution. Unfortunately, he was not paying attention and filled

dalvyx [7]

Answer:

His results will be skewed because there was more water than stock solution. Which would cause the percentage solution to be less than 50% therefore the density would be less than the actual value.

Explanation:

The solution will have percentage less than that of 50%. Therefore the density would be less than the actual value.

Suppose there should be 50 mL of the solution, and he added 60 mL. So 10 mL of the solution is added more.

Suppose the mass of the solute is m.

Originally, the density is = $\frac{m}{50}$ $\left(\frac{\text{mass}}{\text{volume}}\right)$

Now after adding extra 10 mL , the density becomes $\frac{m}{60}$ .

Therefore, $\frac{m}{50}>\frac{m}{60}$

So the density decreases when we add more solution.

4 0

3 years ago

How many grams are in 3.14 x 1015 molecules of CO?

Vesnalui [34]

Answer:

Explanation:

Not Many

1 mol of CO has a mass of

C = 12

O = 16

1 mol = 28 grams.

1 mol of molecules = 6.02 * 10^23

x mol of molecules = 3.14 * 10^15 Cross multiply

6.02*10^23 x = 1 * 3.14 * 10^15 Divide by 6.02*10^23

x = 3.14*10^15 / 6.02*10^23

x = 0.000000005 mols

x = 5*10^-9

1 mol of CO has a mass of 28

5*10^-9 mol of CO has a mass of x Cross Multiply

x = 5 * 10^-9 * 28

x = 1.46 * 10^-7 grams

Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample

5 0

2 years ago

A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a

Semenov [28]

Answer: The metal having molar mass equal to 26.95 g/mol is Aluminium

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = $\frac{1}{2}\times 0.01670=0.00835mol$ of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol$

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

The chemical equation for the reaction of metal (forming $M^{3+}$ ion) and sulfuric acid follows:

$2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2$

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = $\frac{2}{3}\times 0.0556=0.0371mol$ of metal

To calculate the molar mass of metal for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

$0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol$

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

6 0

3 years ago

Which of the following techniques can be used to separate a heterogeneous mixture into its component parts? (3 points)

avanturin [10]

Answer:

Separation by density

Hope this helps

6 0

3 years ago

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