`100 g` of `NaCl` is stirred in `100 mL` of water at `20^(@)C` till the equilibrium is attained: (`a`) How much `NaCl` goes into the solution and how much of it is left undissolved at equilibrium? The solubility of `NaCl` at `20^(@)C` is `6.15 mol//litre`. (`b`) What will be the amount of `NaCl` left undissolved, if the solution is diluted to `200 mL`?

`100 g` of `NaCl` is stirred in `100 mL` of water at `20^(@)C` till th

1.	`100 g` of `NaCl` is stirred in `100 mL` of water at `20^(@)C` till the equilibrium is attained: (`a`) How much `NaCl` goes into the solution and how much of it is left undissolved at equilibrium? The solubility of `NaCl` at `20^(@)C` is `6.15 mol//litre`. (`b`) What will be the amount of `NaCl` left undissolved, if the solution is diluted to `200 mL`?
Answer» Solubility of `NaCl=6.15mol//litre` `=6.15xx58.5g//litre` `=(6.15xx58.5)/(10)g//100mL` `=36g//100mL` (`a`) Thus amount of `NaCl` in `100mL` at `20^(@)C` get dissolved `=36g` and amount of `NaCl` in `100mL` at `20^(@)C` remained undissolved `=100-36=64g` (`b`) If volume of solution is diluted to `200mL`, `36g` more of `NaCl` will be dissolved leaving only `28g NaCl` dissolved in `200 mL`.

Discussion

No Comment Found

Related InterviewSolutions

The vapour pressure of a liquid in a closed container depends uponA. 1 onlyB. 2 onlyC. 1 and 3 onlyD. 1,2,and3
Solubility of a solute in water is dependent on temperature as given by `S=Ae^(-DeltaH//RT)`, where `DeltaH`=heat of solution `"Solute"+H_(2)O(l) hArr "Solution", DeltaH= +- x` For given solution, variation of log S with temperature is shown graphically. Hence, solution is A. `CuSO_(4).5H_(2)O`B. NaClC. SucroseD. CaO
The `K_(c)` for `H_(2(g)) + I_(2(g))hArr2HI_(g)` is 64. If the volume of the container is reduced to one-half of its original volume, the value of the equilibrium constant will beA. `16`B. `32`C. `64`D. `128`
The `K_(c)` for `H_(2(g)) + I_(2(g))hArr2HI_(g)` is 64. If the volume of the container is reduced to one-half of its original volume, the value of the equilibrium constant will beA. 16B. 32C. 64D. 128
Following gaseous reaction is undergoing in a vessel `C_(2)H_(4) + H_(2)hArrC_(2)H_(6)` , `Delta H = - 32.7` Kcal Which will increase the equilibrium concentration of `C_(2)H_(6)` ?A. Increase of temperatureB. By reducing temperatureC. By removing some hydrogenD. By adding some `C_(2)H_(6)`
For the gas phase reaction `C_(2)H_(4)+H_(2) hArr C_(2)H_(6)(DeltaH=-32.7 "kcal")` carried out in a vessel, the equilibrium concentration of `C_(2)H_(4)` can be increased byA. Increasing the temperatureB. increasing concentration of `H_(2)`C. decreasing temperatureD. increasing pressure
The synthesis of ammonia is given as: `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g), DeltaH^(ɵ)=-92.6 kJ mol^(-1)` given `K_(c)=1.2` and temperature `(T)=375^(@)C` On increasing the temperature, the value of equilibrium constant `K_(c)`A. IncreasesB. DecreasesC. Remain unchangedD. Cannot be predicted
In an equilibrium `A+B hArr C+D`, A and B are mixed in vessel at temperature T. The initial concentration of A was twice the initial concentration of B. After the equilibrium has reaches, concentration of C was thrice the equilibrium concentration of B. Calculate `K_(c)`.
For the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g),DeltaH=-93.6 kJ mol^(-1)`. The concentration of `H_(2)` at equilibrium will increase ifA. the temperature is loweredB. the volume of the system is decreasedC. `N_(2)` is added at constant volumeD. `NH_(3)` is added
A tenfold increase in pressure on the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` at equilibrium result in ……….. in `K_(p)`.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment