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Question: Calculate the mass of \(N{a_2}S{O_4}.7{H_2}O\) that contains \(6.02 \times {10^{22}}\) atoms of oxyg...

Calculate the mass of Na2SO4.7H2ON{a_2}S{O_4}.7{H_2}O that contains 6.02×10226.02 \times {10^{22}} atoms of oxygen?

Explanation

Solution

To know the mass of the given compound, Na2SO4.7H2ON{a_2}S{O_4}.7{H_2}O that contains 6.02×10226.02 \times {10^{22}} atoms of oxygen, first we will find the number of moles then we will find the molar mass of the given compound.

Complete step by step answer:
Given Molecular Formula = Na2SO4.7H2ON{a_2}S{O_4}.7{H_2}O
(4+7)\therefore (4 + 7) 0 atoms in each molecule.
Given that-
Number of 0 atoms (N)(N) = 6.02×10226.02 \times {10^{22}}
Number of moles of 0 atoms (n1)({n_1}) =NNA=6.02×10226.02×1023=0.1mole = \dfrac{N}{{{N_A}}} = \dfrac{{6.02 \times {{10}^{22}}}}{{6.02 \times {{10}^{23}}}} = 0.1mole
Now, Moles of Na2SO4.7H2ON{a_2}S{O_4}.7{H_2}O , (n)=n111=0.111=1110mole(n) = \dfrac{{{n_1}}}{{11}} = \dfrac{{0.1}}{{11}} = \dfrac{1}{{110}}mole
So, Molar mass of Na2SO4.7H2ON{a_2}S{O_4}.7{H_2}O (M)=268g/mol(M) = 268g/mol
\therefore Mass of 1110mole\dfrac{1}{{110}}mole
=n×M =1110×268 =2.436g = n \times M \\\ = \dfrac{1}{{110}} \times 268 \\\ = 2.436g \\\

Hence, the mass of Na2SO4.7H2ON{a_2}S{O_4}.7{H_2}O that contains 6.02×10226.02 \times {10^{22}} atoms of oxygen is 2.436g2.436g .

Note: The mass of a given substance (chemical element or chemical compound in g) divided by its amount of substance. Chemists can measure a quantity of matter using mass, but in chemical reactions it is often important to consider the number of atoms of each element present in each sample.