Home
Class 11
CHEMISTRY
0.96 g of HI were, heated to attain equi...

`0.96 g` of HI were, heated to attain equilibrium `2HI hArr H_(2)+I_(2)`. The reaction mixture on titration requires `15.7 mL` of `N//10` hypo solution. Calculate the degree of dissociation of `HI`.

Text Solution

Verified by Experts

The correct Answer is:
`20.9%`
meq of `l_(2)=(W_(l_2))/(127)xx1000=1.57`
`W_(l_(2))=(127xx1.57)/(1000)` `n_(l_(2))=(1.57)/(2000)``2HI(g)harrH_(2)(g)+l_(2)(g)`
`a-aalpha , (aalpha)/(2) , (aalpha)/(2) , (0.96)/(256)xxalpha=(1.57)/(2000)`
`alpha=0.209`
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL EQUILIBRIUM

    RESONANCE ENGLISH|Exercise Board Level Exercise|33 Videos

  • CHEMICAL EQUILIBRIUM

    RESONANCE ENGLISH|Exercise Exercise-1 (Part-1)|38 Videos

  • CHEMICAL EQUILIBRIUM

    RESONANCE ENGLISH|Exercise Solved examples|9 Videos

  • CHEMICAL BONDING

    RESONANCE ENGLISH|Exercise ORGANIC CHEMISTRY(Fundamental Concept )|6 Videos

  • D & F-BLOCK ELEMENTS & THEIR IMPORTANT COMPOUNDS

    RESONANCE ENGLISH|Exercise Match the column|1 Videos

Similar Questions

Explore conceptually related problems

60 mL of H_(2) and 42 mL of I_(2) are heated in a closed vessel. At equilibrium, the vessel contains 20 mL HI. Calculate degree of dissociation of HI.

10 ml of H_(2)O_(2) solution on treatment with Kl and titration of liberated l_(2) required 20 mL of 1 N hypo solution. What is the normality of H_(2)O_(2) solution ?

For 2HI(g) hArr H_2(g)+I_2(g) write the expression of k_c .

In the dissociation of 2HI hArr H_(2)+I_(2) , the degree of dissociation will be affected by

In the equilibrium reaction, 2HI(g)hArr H_(2)(g)+I_(2)(g) , which of the following expressions is true?

In the reversible reaction, 2HI(g) hArr H_(2)(g)+I_(2)(g), K_(p) is

2 mole of H_(2) and "1 mole of "I_(2) are heated in a closed 1 litre vessel. At equilibrium, the vessel contains 0.5 mole HI. The degree of dissociation of H_(2) is

256 g of HI were heated in a sealed bulb at 444°C till the equilibrium was attained. The acid was found to be 22% dissociated at equilibrium. Calculate the equilibrium constant for the reaction 2HI(g) hArr H_2(g) +I_2 (g)

The value of K_(p) for dissociation of 2HI hArr H_(2)+I_(2) is 1.84xx10^(-2) . If the equilibrium concentration of H_(2) is 0.4789 mol L^(-1) , calculate the concentration of HI at equilibrium.

15 mol of H_(2) and 5.2 moles of I_(2) are mixed and allowed to attain eqilibrium at 500^(@)C At equilibrium, the concentration of HI is founf to be 10 mol. The equilibrium constant for the formation of HI is.