Home
Class 11
CHEMISTRY
0.7875 g of crystalline barium hydroxide...

0.7875 g of crystalline barium hydroxide is dissolved in water .For the neutralization of this solution 20 mL of N/4 `HNO_(3)` is required. How many moles of water of crystallization are present in one mole of this base ? (Given : Atomic mass Ba=137,O=16, N=14, H=1)

Text Solution

Verified by Experts

Let the molecular formula be `Ba(OH)_(2). xH_(2)O`
Mol. Mass of `Ba(OH)_(2).xH_(2)O=137.4+(2xx16)+(2xx1)+18x`
=171.4+18x
Eq. mass of `Ba(OH)_(2).xH_(2)O=(171.4+18x)/(2)`
`20 mL(N)/(4)HNO_(3)-=20mL(N)/(4)Ba(OH)_(2).xH_(2)O`
Amount of `Ba(OH)_(2).xH_(2)O=(171.4+18x))/(2xx4)xx(20)/(1000)`
`=(171.4+18x)/(400)g`
Amount of `Ba(OH)_(2).xH_(2)O=0.789 g`
Hence, `(171.4+18x)/(400)=0.789`
or `171.4+18x=0.789xx400`
`x=(144.2)/(18)=8.01~~8`
Thus, 8 g moles of water molecules are present in one g mole of the base.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • VOLUMETRIC ANALYSIS

    OP TANDON|Exercise Example 20|2 Videos

  • VOLUMETRIC ANALYSIS

    OP TANDON|Exercise Example 21|2 Videos

  • VOLUMETRIC ANALYSIS

    OP TANDON|Exercise Example 18|2 Videos

  • THE COLLOIDAL STATE

    OP TANDON|Exercise Self Assessment|21 Videos

Similar Questions

Explore conceptually related problems

14.3 g of Na_(2)CO_(3) xH_(2)O is dissolved in water and the volume is made up to 200 mL. 20 mL of this solution required 40 mL of (N)/(4) HNO_(3) for complete neutralisation. Calculate x

0.5 g of an oxalate was dissolved in water and the solution made to 100 mL. On titration 10 mL of this solution required 15 mL of (N)/(20)KMnO_(4) . Calculate the percentage of oxalate in the sample .

Knowledge Check

  • 0.1914 g of an organic acid is dissolved in about 20 mL of water. 25 mL of 0.12 N NaOH is required for complete neutralization of the acid solution. The equivalent mass of the acid is

    A
    `65.0`
    B
    `64.0`
    C
    `62.5`
    D
    `63.8`

  • 3.92 g of ferrous ammonium sulphate crystals are dissolved in 100 ml of water, 20 ml of this solution requires 18 ml of KMnO_(4) during titration for complete oxidation. The weight of KMnO_(4) present in one litre of the solution is

    A
    `3.476 g`
    B
    `12.38 g`
    C
    (c )` 34.76 g`
    D
    `1.238 g`

  • 0.63 g of diabasic acid was dissolved in water. The volume of the solution was made 100 mL . 20 mL of this acid solution required 10 mL of N//5 NaOH solution. The molecular mass of acid is:

    A
    `63`
    B
    `126`
    C
    `252`
    D
    `128`

    • Similar Questions

    Explore conceptually related problems

    0.25 g of an oxalate salt was dissolved in 100 mL of water. 10 mL of this solution required 8 mL of N//20 KMnO_(4) for its oxidation. Calculate the percentage of oxalate in the salt.

    1.80 g of impure sample of oxalate was dissolved in water and the solution made to 250 mL. On titration 20 mL of this solution required 30 mL of M/50 KMnO_4 solution . Calculated the percentage purity of the sample.

    2.65 g of diacidic base was dissolved in water and made up to 500 mL. 20 of this solution completely neutralised 12 mL of N//6 HCl. Find out the equivalent mass and molecular mass of the base.

    What is the mass of 0.5 mole of water ( H_2O )? (Atomic masses : H = 1 u, O = 16 u).

    2.65gm of a diacidic base was dissolved in 500 ml of water. Twenty millilitres of this solution required 12 ml of (N)/(6) HCl solution. Calculate the equivalent mass and molucular mass of the base.

    Home
    Profile