Question

Find the total number of electrons present in 40 millimoles of $BaS{O}_4.$ (atomic number of Ba = 56)

• A. $2.89\times {10}^{24}$
• B. $3.89\times {10}^{24}$
• C. $2.51\times {10}^{24}$
• D. $3.51\times {10}^{24}$

Answer 1

The correct option is C $2.51\times {10}^{24}$

The total number of electrons present in one molecule of $BaS{O}_4=56+16+32=104$
Therefore, total number of electrons present in one mole of $BaS{O}_4=104\times 6.022\times {10}^{23}$
Hence, total number of electrons present in 40 millimoles of $BaS{O}_4$
$=104\times 6.022\times {10}^{23}\times 40\times {10}^{-3}=2.51\times {10}^{24}$