\left. \begin{array} { l } { ( A ) 6.02 \times 10 ^ { 18 } } \\ { ( C

1.	\left. \begin{array} { l } { ( A ) 6.02 \times 10 ^ { 18 } } \\ { ( C ) 3.0 \times 6.02 \times 10 ^ { 19 } ( 0 ) 3.0 \times 6.02 \times 10 ^ { 18 } } \end{array} \right.
Answer» Barium nitrate is Ba(NO3)2. and in solution.. it will be Ba2+ and 2 NO3- , because both are ions so, no of moles = M.V = 0.001*0.1 = 0.0001moles now, in 1 moles the no. of atoms = 6.02×10²³ so, here the no..of moles and ions is 0.0001×3ions = 0.0003 moles ions so, total no of moles are = 0.0003 × 6.02×10²³ = 3×6.02× 10^19 option C

\left. \begin{array} { l } { ( A ) 6.02 \times 10 ^ { 18 } } \\ { ( C ) 3.0 \times 6.02 \times 10 ^ { 19 } ( 0 ) 3.0 \times 6.02 \times 10 ^ { 18 } } \end{array} \right.

Answer»

Barium nitrate is Ba(NO3)2.

and in solution.. it will be Ba2+ and 2 NO3- , because both are ions

so, no of moles = M.V = 0.001*0.1 = 0.0001moles

now, in 1 moles the no. of atoms = 6.02×10²³

so, here the no..of moles and ions is 0.0001×3ions = 0.0003 moles ions

so, total no of moles are = 0.0003 × 6.02×10²³ = 3×6.02× 10^19

option C