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10 mL of an HCl solution gave 0.1435 g o...

10 mL of an HCl solution gave `0.1435` g of AgCl when treated with excess of `AgNO_(3)` .The normality of the resulting solution is

A

`0.1`

B

3

C

`0.3`

D

`0.2`

Text Solution

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The correct Answer is:
To find the normality of the HCl solution that produced 0.1435 g of AgCl when treated with excess AgNO3, we can follow these steps: ### Step 1: Calculate the moles of AgCl produced To find the moles of AgCl, we use the formula: \[ \text{Moles of AgCl} = \frac{\text{mass of AgCl}}{\text{molar mass of AgCl}} \] The molar mass of AgCl is approximately 143.5 g/mol. \[ \text{Moles of AgCl} = \frac{0.1435 \, \text{g}}{143.5 \, \text{g/mol}} = 0.001 \, \text{mol} \] ### Step 2: Calculate the molarity of HCl Since the reaction between HCl and AgNO3 produces AgCl in a 1:1 molar ratio, the moles of HCl will also be 0.001 mol. Now, we need to convert the volume of HCl from mL to L: \[ \text{Volume of HCl} = 10 \, \text{mL} = 0.010 \, \text{L} \] Now we can calculate the molarity (M) of HCl: \[ \text{Molarity (M)} = \frac{\text{moles of HCl}}{\text{volume of solution in L}} = \frac{0.001 \, \text{mol}}{0.010 \, \text{L}} = 0.1 \, \text{M} \] ### Step 3: Calculate the normality of HCl For HCl, which is a strong acid, the normality (N) is equal to the molarity (M) because it dissociates completely to give one H+ ion per molecule of HCl. \[ \text{Normality (N)} = \text{Molarity (M)} = 0.1 \, \text{N} \] ### Final Answer The normality of the resulting HCl solution is **0.1 N**. ---
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Knowledge Check

  • 0.2435 g of a complex gave 0.2870 g of AgCl when treated with a excess AgNO_(3) solution. The complex is

    A
    `[Cr(NH_(3))_(4)Cl_(2)]Cl`
    B
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    D
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    0.8 M
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