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1 kg of NaOH is added to 10 ml of 0.1 N ...

1 kg of NaOH is added to 10 ml of 0.1 N HCl, the resulting solution will

A

turn blue litmus red

B

turn phenolphthalein solution pink

C

turn methyl orange red

D

will have no effect on red or blue litmus paper

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To solve the problem of determining the resulting solution when 1 kg of NaOH is added to 10 mL of 0.1 N HCl, we can follow these steps: ### Step 1: Calculate the number of equivalents of NaOH To find the number of equivalents of NaOH, we use the formula: \[ \text{Number of equivalents} = \frac{\text{mass (g)}}{\text{equivalent weight (g/equiv)}} \] The equivalent weight of NaOH can be calculated as: \[ \text{Equivalent weight of NaOH} = \frac{\text{Molar mass of NaOH}}{n \text{ factor}} \] Where: - Molar mass of NaOH = 23 (Na) + 16 (O) + 1 (H) = 40 g/mol - Since NaOH is a monoacidic base, its n factor = 1. Now, substituting the values: \[ \text{Equivalent weight of NaOH} = \frac{40 \text{ g/mol}}{1} = 40 \text{ g/equiv} \] Now, substituting this into the number of equivalents formula: \[ \text{Number of equivalents of NaOH} = \frac{1000 \text{ g}}{40 \text{ g/equiv}} = 25 \text{ equivalents} \] ### Step 2: Calculate the number of equivalents of HCl To find the number of equivalents of HCl, we use the formula: \[ \text{Number of equivalents} = \text{Normality} \times \text{Volume (L)} \] First, convert the volume from mL to L: \[ 10 \text{ mL} = 0.01 \text{ L} \] Now, substituting the values: \[ \text{Number of equivalents of HCl} = 0.1 \text{ N} \times 0.01 \text{ L} = 0.001 \text{ equivalents} \] ### Step 3: Compare the number of equivalents Now we compare the number of equivalents of NaOH and HCl: - Equivalents of NaOH = 25 equivalents - Equivalents of HCl = 0.001 equivalents Since the number of equivalents of NaOH (25) is much greater than that of HCl (0.001), the solution will be basic. ### Step 4: Determine the effect on indicators - **Blue litmus paper**: Turns red in acidic solutions (not applicable here). - **Phenolphthalein**: Turns pink in basic solutions (applicable here). - **Methyl orange**: Turns red in acidic solutions (not applicable here). - **No effect on litmus**: Indicates a neutral solution (not applicable here). ### Conclusion The resulting solution will turn phenolphthalein pink, indicating that the solution is basic.

To solve the problem of determining the resulting solution when 1 kg of NaOH is added to 10 mL of 0.1 N HCl, we can follow these steps: ### Step 1: Calculate the number of equivalents of NaOH To find the number of equivalents of NaOH, we use the formula: \[ \text{Number of equivalents} = \frac{\text{mass (g)}}{\text{equivalent weight (g/equiv)}} \] The equivalent weight of NaOH can be calculated as: ...
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NARAYNA-SOLUTIONS & COLLIGATIVE PROPERTIES -EXERCISE : 1 (C.W)
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  2. A solution is obtained by mixing 300 g of 25% solution and 400 g of 40...

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  3. 1 kg of NaOH is added to 10 ml of 0.1 N HCl, the resulting solution wi...

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  4. The solution having lowest molar concentration is

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  6. When an oxide M(2)O(3) is oxidised to M(2)O(5) its equivalent is

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  7. The equivalent weight of Mohr's salt FeSO(4). (NH(4))(2)SO(4). 6H(2)O ...

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  8. Equivalent weight of NH(3) as a base is

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  9. In an oxidation-reduction reaction, MnO(4)^(-) ion is converted to Mn^...

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  10. Number of moles of solute dissolved in 1000 g. of the solvent is calle...

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  12. Which of the following is correct ?

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  13. Which of the following has no unit?

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  14. What is the mole fraction of glucose in 10% w/W glucose solution ?

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  15. Calculate the weight of carbon dioxide dissolved in 1 L bottle of carb...

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  16. The boiling point of C(6)H(6) , CH(3)OH , C(6)H(5)NH(2) "and"C(6)H(5)N...

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  18. Lowering of vapour pressures of equimolar solution of glucose, sodium ...

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  19. Which of the following salt has the same value of Van’t Hoff factor ‘i...

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