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Juliette [100K]
3 years ago
5

Someone give answer and explain please!! (:

Chemistry
1 answer:
MAXImum [283]3 years ago
6 0
<h3><u>Q. No. 5:</u></h3>

               What is the molarity of 3.20 g of Potassium dissolved into 0.50 L solution.

<h3><u>Answer:</u></h3>

                     Molarity is the the unit of concentration and it is expressed as the amount of solute dissolved per unit volume of solution. It is expressed as,

                         Molarity  =  Moles / Volume of Solution  (1)

Data Given;

                  Mass  =  3.20 g

                  Volume  =  0.50 L

                   A.Mass of K  =  39.10 g/mol

First calculate Moles for given mass as,

                   Moles  =  Mass / M.mass

                   Moles  =  3.20 g / 39.10 g.mol⁻¹

                   Moles  =  0.08184 mol

Now, putting value of Moles and Volume in eq. 1,

                        Molarity  =  0.08184 mol ÷ 0.50 L

                        Molarity  =  0.163 mol.L⁻¹ (or) 0.163 M

______________________________________________

<h3><u>Q. No. 6:</u></h3>

               What is the molarity of 2.31 g of Potassium dissolved into 400 mL solution.

<h3><u>Answer:</u></h3>

Data Given;

                  Mass  =  2.31 g

                  Volume  =  400 mL  =  0.40 L

                   A.Mass of K  =  39.10 g/mol

First calculate Moles for given mass as,

                   Moles  =  Mass / M.mass

                   Moles  =  2.31 g / 39.10 g.mol⁻¹

                   Moles  =  0.0590 mol

As,

                   Molarity  =  Moles / Volume of Solution

Now, putting value of Moles and Volume,

                   Molarity  =  0.0590 mol ÷ 0.40 L

                   Molarity  =  0.147 mol.L⁻¹ (or) 0.147 M

______________________________________________

<h3><u>Q. No. 7:</u></h3>

               How many moles of sodium nitrate are there in 75 mL of 0.25 M solution?

<h3><u>Answer:</u></h3>

Data Given;

                  Moles  =  <u>??</u>

                  Volume  =  75 mL  =  0.075 L

                  Molarity  =  0.25 mol.L⁻¹

Molarity is given as,

                        Molarity  =  Moles / Volume of Solution

Solving for Moles,

                        Moles  =  Molarity × Volume of Solution

Putting value of Molarity and Volume,

                        Moles  =  0.25 mol.L⁻¹ × 0.075 L

                        Moles =  0.0187 mol

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Determine the empirical formula of a compound having the following percent composition by mass: K: 24.74%; Mn: 34.76%; O: 40.50%
Dovator [93]

<u>Answer:</u> The empirical formula of the compound is KMnO_4

<u>Explanation:</u>

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

Let the mass of the compound be 100 g

Given values:

% of K = 24.7%

% of Mn = 34.76%

% of O = 40.50%

Mass of K = 24.7 g

Mass of Mn = 34.76 g

Mass of O = 40.50 g

To calculate the empirical formula of a compound, few steps need to be followed:

  • <u>Step 1:</u> Calculating the number of moles of each element

We know:

Molar mass of K = 39.10 g/mol

Molar mass of Mn = 54.94 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of K}=\frac{24.7g}{39.10g/mol}=0.632 mol

\text{Moles of Mn}=\frac{34.76g}{54.94g/mol}=0.633 mol

\text{Moles of O}=\frac{40.50g}{16g/mol}=2.53 mol

  • <u>Step 2:</u> Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.632 moles

\text{Mole fraction of K}=\frac{0.632}{0.632}=1

\text{Mole fraction of Mn}=\frac{0.633}{0.632}=1

\text{Mole fraction of O}=\frac{2.53}{0.632}=4

  • <u>Step 3:</u> Writing the mole fraction as the subscripts of each of the element

The empirical formula of the compound becomes K_1Mn_1O_4=KMnO_4

Hence, the empirical formula of the compound is KMnO_4

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Molar mass of C2H6O2 = (12 x 2) + (1 x 6) + (2 x 16)

                                        = 62

mole = mass/molar mass

Moles of C2H6O2 = 20/62

                               = 0.322 mol

Moles of water = 80/18

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mole fraction = mol of solute / tatal mol of solution

Mole fraction of ethylene glycol = 0.322/(0.322 + 4.444)

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Mole fraction of water = 1 - 0.068

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