Question
Match List-I with List-II.
| List-I Mass of substance |
List-II Number of atoms |
|---|---|
| A. $\ 1.8mg$ water | I. $2 \times 10^{- 4} \times N_{A}$ |
| B. $\ 9.8mg$ sulphuric acid | II. $\ 1.5 \times 10^{- 4} \times N_{A}$ |
| C. $\ 1.8mg$ carbon | III. $3 \times 10^{- 4} \times N_{A}$ |
| D. 5.85 mg salt $(NaCl)$ | IV. $\ 7 \times 10^{- 4} \times N_{A}$ |
List-II
Number of atoms
Choose the correct answer from the options given below:
Options
Solution
Sol. [Key Concept: Number of atoms $= \frac{mass}{M} \times \ N_{A} \times (atomicity)$; water gives $3 \times 10^{- 4}N_{A}$, $H_{2}{SO}_{4}$ gives $7 \times 10^{- 4}N_{A}$, carbon $1.5 \times 10^{- 4}N_{A}$ and NaCl $2 \times 10^{- 4}N_{A}$, so A-III, B-IV, C-II, D-I.]
(A) No. of atoms in 1.8 mg water $= \frac{1.8 \times 10^{- 3}}{18} \times N_{A} \times 3 = 3 \times 10^{- 4}{\text{ }N}_{A}$
(B) No. of atoms in $9.8mgH_{2}{SO}_{4}$ $= \frac{9.8 \times 10^{- 3}}{98} \times N_{A} \times 7 = 7 \times 10^{- 4} \times N_{A}$
(C) No. of atoms in 1.8 mg carbon $= \frac{1.8 \times 10^{- 3}}{12} \times N_{A} = 1.5 \times 10^{- 4} \times N_{A}$
(D) No. of atoms in 5.85 mg NaCl $= \frac{5.85 \times 10^{- 3}}{58.5} \times N_{A} \times 2 = 2 \times 10^{- 4}{\text{ }N}_{A}$
Correct match : A-III, B-IV, C-II, D-I
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