Calculate the molarity of each of the following solutions. part a) 0.12 mol of lino3 in 5.5 l of solution part b) 60.7 g c2h6o in 2.48 l of solution part c) 14.2 mg ki in 100 ml of solution

For A: The molarity of solution is 0.218 M.

For B: The molarity of solution is 0.532 M.

For C: The molarity of solution is

Explanation:

Molarity is defined as the number of moles present in one liter of solution.

Mathematically,

Or,

We are given:

Moles of = 0.12 moles

Volume of the solution = 5.5 L

Putting values in above equation, we get:

Hence, the molarity of solution is 0.0218 M.

We are given:

Given mass of = 60.7 g

Molar mass of = 46 g/mol

Volume of the solution = 2.48 L

Putting values in above equation, we get:

Hence, the molarity of solution is 0.532 M.

We are given:

Given mass of KI = 14.2 mg =     (Conversion factor:

Molar mass of KI = 166 g/mol

Volume of the solution = 100 L

Putting values in above equation, we get:

Hence, the molarity of solution is

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).