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Chemistry · Some Basic Concepts of Chemistry
हिन्दीMatch List-I with List-II: List-I List-II A Mass per cent I \( \frac{\text{Moles of solute}}{\text{Volume of solution in L}} \) B Molarity II \( \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100 \) C Molality III \( \frac{n_A}{n_A + n_B} \) D Mole fraction IV \( \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \) Choose the correct answer from the options given below:
Match List-I with List-II:
| List-I | List-II | ||
|---|---|---|---|
| A | Mass per cent | I | \( \frac{\text{Moles of solute}}{\text{Volume of solution in L}} \) |
| B | Molarity | II | \( \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100 \) |
| C | Molality | III | \( \frac{n_A}{n_A + n_B} \) |
| D | Mole fraction | IV | \( \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \) |
Choose the correct answer from the options given below:
Correct answer · Option 2
A-(II), B-(I), C-(IV), D-(III)
Explanation
Correct matching is:
- Mass per cent (A) = \(\frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100\) (II)
- Molarity (B) = \(\frac{\text{Moles of solute}}{\text{Volume of solution in L}}\) (I)
- Molality (C) = \(\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}\) (IV)
- Mole fraction (D) = \(\frac{n_A}{n_A + n_B}\) (III)
Therefore, A-(II), B-(I), C-(IV), D-(III).
