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Oxana [17]
2 years ago
8

26.6 mL of 2.50 M stock solution of sucrose is diluted to 50.0 mL. A 16.0 mL sample of the resulting solution is then diluted to

45.0 mL. A 20.0 mL sample of this solution is then diluted to 75.0 mL. What is the concentration of sucrose in the final solution?
Chemistry
1 answer:
frozen [14]2 years ago
7 0

Answer:

In the final solution, the concentration of sucrose is 0.126 M

Explanation:

Hi there!

The number of moles of solute in the volume taken from the more concentrated solution will be equal to the number of moles of solute in the diluted solution. Then, the concentration of the first solution can be calculated using the following equation:

Ci · Vi = Cf · Vf

Where:

Ci = concentration of the original solution

Vi = volume of the solution taken to prepare the more diluted solution.

Cf = concentration of the more diluted solution.

Vf = volume of the more diluted solution.

For the first dillution:

26.6 ml · 2.50 M = 50.0 ml · Cf

Cf = 26.6 ml · 2.50 M / 50.0 ml

Cf = 1.33 M

For the second dilution:

16.0 ml · 1.33 M = 45.0 ml · Cf

Cf = 16.0 ml · 1.33 M / 45.0 ml

Cf = 0.473 M

For the third dilution:

20.0 ml · 0.473 M = 75.0 ml · Cf

Cf = 20.0 ml · 0.473 M / 75.0 ml

Cf = 0.126 M

In the final solution, the concentration of sucrose is 0.126 M

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<u>Answer:</u> The molecular formula for the menthol is C_{10}H_{20}O

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=0.2829g

Mass of H_2O=0.1159g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.2829 g of carbon dioxide, \frac{12}{44}\times 0.2829=0.077g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.1159 g of water, \frac{2}{18}\times 0.1159=0.0129g of hydrogen will be contained.

  • Mass of oxygen in the compound = (0.1005) - (0.077 + 0.0129) = 0.0106 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0129g}{1g/mole}=0.0129moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0106g}{16g/mole}=0.00066moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00066 moles.

For Carbon = \frac{0.0064}{0.00066}=9.69\approx 10

For Hydrogen  = \frac{0.0129}{0.00064}=19.54\approx 20

For Oxygen  = \frac{0.00066}{0.00066}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

Hence, the empirical formula for the given compound is C_{10}H_{20}O_1=C_{10}H_{20}O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 156.27 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

n=\frac{156.27g/mol}{156g/mol}=1

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O

Thus, the molecular formula for the menthol is C_{10}H_{20}O

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