Question

1.61 gm of $N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$ contains same number of oxygen atoms as present in x gm of ${{H}_{2}}S{{O}_{4}}$. The value of x is:
A) 1.28
B) 1.38
C) 1.58
D) 1.78

Answer 1

The answer in obtained based on the calculation of number of oxygen present with the help of the formula which relates the mass and molar mass and also the Avogadro constant and the number of oxygen atom present in $N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$

Complete step by step answer:
- The concept of inorganic chemistry which deals with the calculation of number of moles of the substance present, number of molecules present and also about the number of particular atoms present are familiar to us.
Now, let us focus on calculating the number of oxygen atoms present in sulphuring acid that is the value of x.
- Now, the number of oxygen atoms in the compound can be found by using the formula,
Number of oxygen atom = $\dfrac{m}{M}\times {{N}_{A}}\times$oxygen atom in $N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$
where m is the given mass of the substance.
M is the molar mass of the substance
${{N}_{A}}$ is the Avogadro constant
- Now, molar mass of the substance will be $46 + 32 + 64 + \left( 18\times 10 \right) = 322g/mol$
Now, by substituting the values accordingly from the given data,
Number of oxygen atoms$=\dfrac{1.61}{322}\times {{N}_{A}}\left( 4+10 \right)$
$\Rightarrow$Number of oxygen atoms = $0.07{{N}_{A}}$
Now, let us consider each options given and obtain the value of x
The general formula for this will be,
Number of oxygen atom = oxygen atoms in ${{H}_{2}}S{{O}_{4}}\times \dfrac{m}{M}\times {{N}_{A}}$
A) In 1.28 g${{H}_{2}}S{{O}_{4}}$, the number of oxygen atom will be,
$4\times \dfrac{1.28}{98}\times {{N}_{A}} = 0.052{{N}_{A}}$
B) In 1.38 g${{H}_{2}}S{{O}_{4}}$, the number of oxygen atom will be,
$4\times \dfrac{1.38}{98}\times {{N}_{A}} = 0.056{{N}_{A}}$
C) In 1.58 g${{H}_{2}}S{{O}_{4}}$, the number of oxygen atom will be,
$4\times \dfrac{1.58}{98}\times {{N}_{A}} = 0.064{{N}_{A}}$
D) In 1.78 g${{H}_{2}}S{{O}_{4}}$, the number of oxygen atom will be,
$4\times \dfrac{1.78}{98}\times {{N}_{A}} = 0.072{{N}_{A}}\approx 0.07{{N}_{A}}$
The correct option is option “D” .

Note: Note that one mole of any substance contains Avogadro's number of molecules that is the Avogadro constant value which is equal to $6.022\times {{10}^{23}}$ number of molecules in it and this value is used in conversion from grams to moles followed by number of molecules.