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Volume of 0.1M HCI required to react com...

Volume of `0.1M HCI` required to react completely with 1g equimolar mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` is

A

`100mL`

B

`200mL`

C

`157.9mL`

D

`159.7 mL`

Text Solution

Verified by Experts

The correct Answer is:
C
`NaCO_(3) +2HCI rarr 2NaCI +H_(2)O +CO_(2)`
`NaHCO_(3)+HCI rarr NaCI +H_(2)O +CO_(2)`
If the mixture contains 1 mol of each. `HCI` required is `(2+1) = 3` mol
Molar mass of `Na_(2)CO_(3) = 106 g mol^(-1)`
Molar mass of `NaHCO_(3) = 84 g mol^(-1)`
Thus, `190g (106 g +84g)` of mixture require `HCI = 3` mol
1g of mixture require `HCI =(3)/(190)` mol
`:.` Vol of `0.M HCI` which contains `(3//190)` mol
`= (1000 xx 3)/(190 xx 0.1) = 157.89mL`
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Knowledge Check

  • The volume of 0.1 M HCl required to neutralise completely 2 g of an equimolar mixture of Na_(2)CO_(3)andNaHCO_(3) is

    A
    318.76 mL
    B
    325 mL
    C
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    D
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    A
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    B
    `152.6` ml
    C
    `200` ml
    D
    `98.5` ml

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    B
    510.5 mg
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