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Calculate the molarity of each of the following solutions:
(a) 0.195 g of cholestrol, C27H46O, in 0.100 L of serum, the average concentration of cholestrol in human serum
(b) 4.25 gram of NH3 in 0.500 L solution, the concentration of NH3 in household ammonia
(c) 1.49 kg of isopropyl alcohol C3H7OH in 2.50 L of solution the concentration of isopropyl alcohol in rubbing alcohol
(d)0.029 gram of I2 in 0.100 L of solution the solubility of I2 in water at 20 C

Respuesta :

Answer:

For a: The molarity of solution is 0.005 M.

For b: The molarity of solution is 0.5 M.

For c: The molarity of solution is 9.92 M.

For d: The molarity of solution is 0.0011 M.

Explanation:

To calculate the molarity of solution, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]  .....(1)

  • For a:

Mass of cholestrol = 0.195 g

Molar mass of cholestrol = 386.65 g/mol

Volume of solution = 0.100 L

Putting values in equation 1, we get:

[tex]\text{Molarity of cholestrol}=\frac{0.195g}{386.65g/mol\times 0.1L}\\\\\text{Molarity of cholestrol}=0.005M[/tex]

Hence, the molarity of solution is 0.005 M.

  • For b:

Mass of ammonia = 4.25 g

Molar mass of ammonia = 17 g/mol

Volume of solution = 0.500 L

Putting values in equation 1, we get:

[tex]\text{Molarity of ammonia}=\frac{4.25g}{17g/mol\times 0.5L}\\\\\text{Molarity of ammonia}=0.5M[/tex]

Hence, the molarity of solution is 0.5 M.

  • For c:

Mass of isopropyl alcohol = 1.49 kg = 1490 g  (Conversion factor: 1 kg = 1000 g)  

Molar mass of isopropyl alcohol = 60.1 g/mol

Volume of solution = 2.5 L    

Putting values in equation 1, we get:

[tex]\text{Molarity of isopropyl alcohol}=\frac{1490g}{60.1g/mol\times 2.5L}\\\\\text{Molarity of isopropyl alcohol}=9.92M[/tex]

Hence, the molarity of solution is 9.92 M.

  • For d:

Mass of iodine = 0.029 g

Molar mass of iodine = 253.8 g/mol

Volume of solution = 0.100 L

Putting values in equation 1, we get:

[tex]\text{Molarity of iodine}=\frac{0.029g}{253.8g/mol\times 0.1L}\\\\\text{Molarity of iodine}=0.0011M[/tex]

Hence, the molarity of solution is 0.0011 M.

The study of chemicals and bonds is called chemistry. There are two types of elements and these are metals and metals.

The correct answer is mentioned below.

What is morality?

  • Molar concentration is a measure of the concentration of a chemical species.
  • in particular of a solute in a solution, in terms of the amount of substance per unit volume of solution

The data is given as follows:-

  • Mass of cholesterol = 0.195 g
  • Molar mass of cholesterol = 386.65 g/mole
  • Volume of solution = 0.100 L

The molarity of the solution is as follows:-

[tex]M = \frac{0.195}{386.65} \\=0.005M[/tex]

Hence, the molarity of the solution is 0.005 M.

For b:

  • Mass of ammonia = 4.25 g
  • Molar mass of ammonia = 17 g/mol
  • Volume of solution = 0.500 L

The molarity of the solution is as follows:-

[tex]M = \frac{4.25}{17}\\0.5M[/tex]

Hence, the molarity of the solution is 0.5 M.

  • Mass of isopropyl alcohol = 1.49 kg = 1490 g  (Conversion factor: 1 kg = 1000 g)  
  • Molar mass of isopropyl alcohol = 60.1 g/mol
  • Volume of solution = 2.5 L    

The molarity will be:-

[tex]M =\frac{1490}{60.1} \\\\=9.92[/tex]

Hence, the molarity of the solution is 9.92 M.

For d:

  • Mass of iodine = 0.029 g
  • Molar mass of iodine = 253.8 g/mol
  • Volume of solution = 0.100 L

[tex]M= \frac{0.029}{253.9} \\=0.0011[/tex]

Hence, the molarity of the solution is 0.0011 M.

For more information about the molarity, refer to the link:-

https://brainly.com/question/14688752

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