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What volume (in ml) of 0.8M AlCl(3) solu...

What volume (in ml) of 0.8M `AlCl_(3)` solution should be mixed with 50 ml of 0.2 M `CaCl_(2)` solution to get solution of chloride ion concentration equal to 0.6 M ?

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The correct Answer is:
5.56 ml
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Knowledge Check

  • 20 ml of CaCl_(2) solution is mixed with 30 ml of 0.1MAlCl_(3) solution. If the resulting solution has chloride ion concentration equal is 0.34 M, the concentration of CaCl_(2) solution added is :

    A
    `0.2` M
    B
    `0.7` M
    C
    `0.4` M
    D
    `0.1` M

  • 20 ml of 0.2M Al_2(SO_4)_3 is mixed with 20 ml of 6.6 M BaCl_2 the concentration of Cl^- ion in solution is

    A
    0.2M
    B
    6.6M
    C
    0.02 M
    D
    0.06 M

  • 100 ml of 0.3 M HCl solution is mixed with 200 ml of 0.3 M H_(2)SO_(4) solution. What is the molariyt of H^(+) in resultant solution ?

    A
    `0.9`
    B
    `0.6`
    C
    `0.4`
    D
    `0.5`

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    500 ml of 0.1 M AlCl_(3) is mixed with 500 ml of 0.1 M MgCl_(2) solution. Then calculation the molarity of Cl^(-) in final solution.

    If 200ml of 0.2 M BaCl_(2) solution is mixed with 500ml of 0.1M Na_(2)SO_(4) solution. Calculate osmotic pressure of resulting solutions?

    50mL of "3MHCl" solution is mixed with 150mL of 1M FeCl_(3) solution.The concentration of chloride ion in the resultant solution is (in mole/lit)

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    What volume of 0.00300 M Hcl solution would just neutralize 30.0 mL of 0.00100 M Ca(OH)_(2) solution?

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