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

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:

 .....(1)

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:

Hence, the molarity of solution is 0.005 M.

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:

Hence, the molarity of 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    

Putting values in equation 1, we get:

Hence, the molarity of solution is 9.92 M.

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:

Hence, the molarity of solution is 0.0011 M.