If the total mass of oxygen present in $F e S O_{4} .7 H_{2} O$ is $176 g$ the total number of electrons in $F e S O_{4} .7 H_{2} O$ is y $\times 10^{25}$ , then the value of y is

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Solution

We know, $176 g$ of O atom = $\frac{176}{16}$ mol of $O$ atoms
= 11 mol of $O$ atoms
Again, from the formula unit of, $F e S O_{4} .7 H_{2} O$
11 $O$ atoms are present in one molecule of $F e S O_{4} .7 H_{2} O$
So, 11 mol of $O$ atoms are present in 1 mol of $F e S O_{4} .7 H_{2} O$
Number of electrons in one formula unit of $F e S O_{4} .7 H_{2} O = 26 + 16 + 11 \times 8 + 14 \times 1 = 144$
Therefore, number of electrons in 1 mol of $F e S O_{4} .7 H_{2} O$ $= 144 \times 6.022 \times 10^{23}$
$= 8.67 \times 10^{25}$

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If the total mass of oxygen present in FeSO4.7H2O is 176 g the total number of electrons in FeSO4.7H2O is y ×1025, then the value of y is

If the total mass of oxygen present in $F e S O_{4} .7 H_{2} O$ is $176 g$ the total number of electrons in $F e S O_{4} .7 H_{2} O$ is y $\times 10^{25}$ , then the value of y is