Q1 / 50 · Manipal (MET) 2023 · Chemistry

The number of $\mathrm{Cl}^{-}$ions in 100 mL of 0.001 M NaCl solution is

  • A
    $6.022 \times 10^{23}$
  • B
    $6.022 \times 10^{20}$
  • Correct answer
    C
    $6.022 \times 10^{19}$
  • D
    $6.022 \times 10^{21}$

Solution

100 mL of $0.001 \mathrm{M} \mathrm{NaCl}^{-2}$ contain $\mathrm{Cl}^{-1}$ ions $\begin{aligned} & =\frac{0.001}{1000} \times 100=10^{-4} \mathrm{~mol} \\ & =10^{-4} \times\left(6.022 \times 10^{23}\right)=6.022 \times 10^{19} \mathrm{Cl}^{-} \text {ions }\end{aligned}$

Hence, the correct answer is option: 'C'.

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