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The molarity of solution obtained by dis...

The molarity of solution obtained by dissolving 0.01 moles of NaCl in 500 ml of solution is

A

0.01 M

B

0.005 M

C

0.02 M

D

0.1 M

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The correct Answer is:
To find the molarity of the solution obtained by dissolving 0.01 moles of NaCl in 500 ml of solution, we can follow these steps: ### Step 1: Understand the formula for molarity Molarity (M) is defined as the number of moles of solute divided by the volume of the solution in liters. The formula is: \[ \text{Molarity (M)} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \] ### Step 2: Identify the number of moles of solute From the question, we know that the number of moles of NaCl (solute) is: \[ \text{Number of moles of NaCl} = 0.01 \text{ moles} \] ### Step 3: Convert the volume of the solution from milliliters to liters The volume of the solution is given as 500 ml. To convert this to liters, we use the conversion factor that 1 liter = 1000 ml: \[ \text{Volume in liters} = \frac{500 \text{ ml}}{1000} = 0.5 \text{ liters} \] ### Step 4: Substitute the values into the molarity formula Now, we can substitute the values we have into the molarity formula: \[ \text{Molarity (M)} = \frac{0.01 \text{ moles}}{0.5 \text{ liters}} \] ### Step 5: Calculate the molarity Now, we perform the calculation: \[ \text{Molarity (M)} = \frac{0.01}{0.5} = 0.02 \text{ M} \] ### Final Answer The molarity of the solution is: \[ 0.02 \text{ M} \] ---

To find the molarity of the solution obtained by dissolving 0.01 moles of NaCl in 500 ml of solution, we can follow these steps: ### Step 1: Understand the formula for molarity Molarity (M) is defined as the number of moles of solute divided by the volume of the solution in liters. The formula is: \[ \text{Molarity (M)} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \] ### Step 2: Identify the number of moles of solute From the question, we know that the number of moles of NaCl (solute) is: ...
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NARAYNA-SOLUTIONS & COLLIGATIVE PROPERTIES -EXERCISE : 2 (H.W)
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  3. The molarity of solution obtained by dissolving 0.01 moles of NaCl in ...

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  4. The weight in grams of KCl (Mol.wt. = 74.5) in 100 ml of a 0.1 M KCl s...

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  5. The weight of AgNO(3) (Mol.wt = 170) present in 100 ml of 0.25 M solut...

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  6. Which of the following should be done in order to prepare 0.40 M NaCI ...

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  7. 250 ml of a solution contains 6.3 grams of oxalic acid (mol.wt. = 126)...

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  8. The volume of 0.25 MH(3)PO(3) required to neutralise 25 ml of 0.03 M C...

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  9. How many grams of a dibasic acid (Mol. Wt. =200) should be present in ...

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  11. ml of 0.1 M H(2)SO(4) is required to neutralise 50 ml of 0.2 M NaOH so...

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  12. If 20 mL of 0.4 N NaOH solution completely neutralises 40 mL of a diba...

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  13. Choose the correct answer among the following is (i) Normality and...

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  14. Calculate the amount of benzoic acid (C(6)H(5)COOH) required for prepa...

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  15. Concentrated nitric acid used for laboratory works is 68% nitric acid ...

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  16. A solution of glucose in water is labelled as 10% w/w. What would be t...

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  17. Calculate molality of 2.5g of ethanoic acid (CH(3)COOH) in 75 g of ben...

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  18. X grams of water is mixed in 69 g of ethanol. Mole fraction of ethanol...

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  19. A gas mixture 44g of CO(2) and 14g of N(2), what will be fraction of C...

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