Question

Calculate the number of ba2+ ions and cl- ions present in 104.1g of anhydrous bacl2

Answer 1

Answer:Number $Ba^{2+}$ ions:=$3.0103\times 10^{23} ions$

Number of $Cl^-$ ions:=$6.0106\times 10^{23} ions$

Explanation:

Mass of barium chloride =  104.1 g

Mole of barium chloride = $\frac{104.1 g}{208.23 g/mol}=0.4999 mol$

Number molecules of barium chloride =$0.4999 \times 6.022\times 10^{23]=3.0103\times 10^{23} molecules$

Number $Ba^{2+}$ ions:

$1\times 3.0103\times 10^{23} ions=3.0103\times 10^{23} ions$

Number of $Cl^-$ ions:

$2\times 3.0103\times 10^{23} ions=6.0106\times 10^{23} ions$

Answer 2

Number of Ba ions=3.01*10^23 ions

Number of CI ions=6.01*10^23 ions