The number of ions present in 1 ml of 0.1 M CaCl2 solution is

Molarity of CaCl2 = 0.1M Volume of CaCl2 = 1ml Total number of ions = = = =

The number of ions present in 1 ml of 0.1 M CaCl2 solution is

Molarity of CaCl2 = 0.1M Volume of CaCl2 = 1ml Total number of ions = = = =

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The number of ions present in 1 ml of 0.1 M CaCl₂ solution is

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1.8 $\times$ 10²⁰

6.0 $\times$ 10²⁰

1.8 $\times$ 10¹⁹

1.8 $\times$ 10²¹

answer is A.

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Detailed Solution

Molarity of CaCl₂ = 0.1M

Volume of CaCl₂ = 1ml

$large {M_{CaC{l_2}}} = {n_{CaC{l_2}}} times frac{{1000}}{{{V_{ml}}}}$

$large 0.1 = {n_{CaC{l_2}}} times frac{{1000}}{1}$

$large {n_{CaC{l_2}}} = 10^{-4}$

$large mathop {CaC{l_2}} limits_{{{10}^{ - 4}}mole}^{mole}rightarrow mathop {C{a^{ + 2}}}limits_{{{10}^{ - 4}}{N_A}}^{{N_A}} + mathop {2C{l^ - }}limits_{2 times {{10}^{ - 4}}{N_A}}^{2{N_A}}$

Total number of ions = $large 3 times 10^{-4} ions$

= $large 3 times 10^{-4} times 6.023 times 10^{23}$

= $large 18 times 10^{19}$

= $large 18 times 10^{20}$

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The number of ions present in 1 ml of 0.1 M CaCl₂ solution is