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22.6 g of an ammonium salt were treated ...

22.6 g of an ammonium salt were treated with 100 mL of normal NaOH solution and boiled till no more of ammonia gas was given off. The excess of NaOH solution left over required 60 mL normal sulphuric acid. Calculate the percentage of ammonia in the salt.

Text Solution

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60 mL normal `H_(2)SO_(4)-=60 mL` normal NaOH
Thus, (100-60) mL normal NaOH were consumed by ammonium salt.
So, 40 mL normal `NaOH -= 40 mL` normal `NH_(3)`
Amount of `NH_(3)` in 40 mL normal `NH_(3)`
`=("Eq. mass of " NH_(3)xx40)/(1000)`
`=(17xx40)/(1000)=0.68`
SO, % of ammonia in the ammonium salt `=(0.68)/(2.26)xx100`
=30.09
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Knowledge Check

  • 2.20 g of an ammonium salt was boild with 74 ml of 1 N NaOH till the emission of ammonia gas ceased. The excess of unused NaOH solution required 70 ml of N/2 sulphuric acid for neutralization. The percentage of ammonia in the salt is:

    A
    0.31
    B
    0.83
    C
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    D
    0.47

  • The density of a solution containing 13% by mass of sulphuric acid is 1.09g//mL . Calculate the molarity and normality of the solution

    A
    ` 1.445M `
    B
    ` 14.45M `
    C
    ` 144.5M `
    D
    ` 0.1445M `

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