Question

Calculate the number of each atom present in $106.5$ g of $NaCl{O}_3$.

• A. $6.023\times {10}^{23}$ atoms of Na, $18.06\times {10}^{23}$ atoms of Cl, $18.06\times {10}^{23}$ atoms of O
• B. $6.023\times {10}^{23}$ atoms of Na, $6.023\times {10}^{23}$ atoms of Cl, $6.023\times {10}^{23}$ atoms of O
• C. $6.023\times {10}^{23}$ atoms of Na, $6.023\times {10}^{23}$ atoms of Cl, $18.06\times {10}^{23}$ atoms of O
• D. none of these

Answer 1

The correct option is C $6.023\times {10}^{23}$ atoms of Na, $6.023\times {10}^{23}$ atoms of Cl, $18.06\times {10}^{23}$ atoms of O

Molar mass of $NaCl{O}_3=106.5gm$. So, $1$ mole of it contains $6.023\times {10}^{23}$ atom Na, $6.023\times {10}^{23}$ atom Cl, $18.06\times {10}^{23}$ atom O.