Question

Four beakers containing potassium nitrate dissolved in water are allowed to evaporate to dryness. Beakers 1 through 4 contain 2.3, 1.91, 5.985, and 0.52 g of dry potassium nitrate respectively. How many moles of potassium nitrate were recovered after the water evaporated?

Answer 1

Explanation:

Molar mass of potassium nitrate will be calculated as follows.

   Molar mass $KNO_{3}$ = molar mass of K + molar mass of N + 3 × molar mass of O

                             = 39.098 g/mol + 14.006 g/mol + 3 × 15.999 g/mol

                             = 102.102 g/mol

Now, adding the given amount of potassium nitrate present in each beaker as follows.

               (2.3 + 1.91 + 5.985 + 0.52) g

               = 10.715 g

Therefore, calculate number of moles as follows.

                   No. of moles = $\frac{mass}{\text{molar mass}}$

                                         = $\frac{10.715 g}{102.102 g/mol}$

                                         = 0.105 mol

Thus, we can conclude that 0.105 moles of potassium nitrate were recovered after the water evaporated.