Question

Calculate the molarity of each of the following solutions (a) 30g of \[Co{{\left( N{{O}_{3}} \right)}_{2}}.6{{H}_{2}}O\]in 4.3L of solution (b) 30 mL of 0.5M ${{H}_{2}}S{{O}_{4}}$diluted to 500 mL.

Answer 1

Hint: The answer includes the formula used for the calculation of molarity given by M=$\dfrac{{{n}_{solute}}}{{{v}_{solution}}}$ and to find the molar mass of the given complex and an acid which leads to the molarity of the solutions.

Complete answer:
We know from the chapters of chemistry that the molar concentration or molarity is the measure of concentration of chemical species which in particular is the solute per unit volume of the solution.
Thus, we can write molarity as,
M=$\dfrac{{{n}_{solute}}}{{{v}_{solution}}}$
We know that the molar mass of the complex or solute is calculated as,
(a) Molar mass of the complex \[Co{{\left( N{{O}_{3}} \right)}_{2}}.6{{H}_{2}}O\]= 58.93+{[14+(16$\times $3)]$\times $2 }+ 18$\times $6 = 291.04
Thus, molar mass of complex=291.04$\approx $291 g/ mol.
Also, number of moles of\[Co{{\left( N{{O}_{3}} \right)}_{2}}.6{{H}_{2}}O\]present in 30 g is given by,
Moles of \[Co{{\left( N{{O}_{3}} \right)}_{2}}.6{{H}_{2}}O\]= $\dfrac{30}{291}$ = 0.103 mol.
Now, molarity of the complex in the solution is given by,
  Molarity = $\dfrac{0.103}{4.3L}$ = 0.023 M
Therefore molarity of complex \[Co{{\left( N{{O}_{3}} \right)}_{2}}.6{{H}_{2}}O\]in solution is 0.023 M.
(b) Next, we shall calculate molarity similarly for ${{H}_{2}}S{{O}_{4}}$which is diluted to 500 mL.
We know that dilution of acid is done using water.
Therefore, 500 mL refers to 0.5 M solution having 0.5 mol present in 1000 mL of acid solution.
Now, total number of moles present in 30 mL of the acid solution i.e., 0.5 M ${{H}_{2}}S{{O}_{4}}$will be equal to $\dfrac{0.5\times 30}{1000}$ .
Simplifying this problem we get 0.015 mol.
Thus, the molarity of the solution is given by’
Molarity=$\dfrac{0.015}{0.5L}$ = 0.03 M
Therefore, molarity of ${{H}_{2}}S{{O}_{4}}$in 30 mL of solution = 0.03 M

Note:
While calculation of molarity of any solution note that the volume written is in the form of litres of ‘solution’ and not in terms of litres of ‘solvent’. During the final reporting of the answer make sure that the molarity unit is written as ‘M’ and read as ‘molar’.