When `100mL` of `1.0M HCl` was mixed with `100 mL` of `1.0 M NaOH` in an insulated beaker at constant pressure, a temperature increase of `5.7^(@)C` w

Monjur Alahi

When `100mL` of `1.0M HCl` was mixed with `100 mL` of `1.0 M NaOH` in an insulated beaker at constant pressure, a temperature increase of `5.7^(@)C` was measured for the beaker and its contents (Expt.1). Because the enthalpy of neutralisation of a strong acid with a strong base is constant `(-57.0kJmol^(-1))` this experiment couldbe used to measure the calorimeter constant. In a second experiment (Expt.2), `100mL` of `2.0M` acetic acid `K_(a)=2.0xx10^(-5))` was mixed with `100mL` of `1.0M NaOH` (under identical conditions to Expt. 1) where a temperature rise of `5.6^(@)C` was measured.
Enthalpy of dissociation (in `kJ mol^(-1)`) of acetic acid obtained from the Expt. 2 is
A. 1
B. 10
C. 24.5
D. 51.4


0 like 0 dislike
1 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

Correct Answer - A
Number of moles of HCl and NaOH
`=(MV)/(1000)=(1xx100)/(1000)=0.1`
Heat released in neutralization `=0.1xx57=5.7kJ`
`=5700kJ`
Heat used to increase the temperature of mixed solution
`=msDeltaT`
`=200xx4.2xx5.7`
`=4788J`
Heat used to increase the temperature of calorimeter
`=5700-4788=912J`
i.e., `msDeltaT=912`
`msxx5.7=912`
`ms=160JK^(-1)` for calorimeter
Heat released in neutralization of `CH_(3)COOH and NaOH`
`=(msDeltaT)_("solution")+(msDeltaT)_("calorimeter")`
`=(200xx4.2xx5.6)+(160+5.6)=5600J`
Heat used in the dissociation of `CH_(3)COOH` per mol
`=(5700-5600)/(0.1)`
`=1000J=1kJ" "mol^(-1)`.

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
One litre of `N//2` HCl solution is heated in a beaker and it is onbserved that when volume was reduced to 600 mL, 3.25 g of HCl is lost. Calculate th

Normality of solution=`("Massof HCI"//"Equivalent mass")/("Volume of solution in litres")` `(0.5 equiv.L^(-1))=(Mass of HCI)/((36.5 g equiv^(-1))xx(1L))` Mass of HCI left= 18.25-3.25=15.0g Volume of resulting solution =`600mL600/1000=0.6L` Normality (N)= `("Mas of acid"//"Gram...

2 Answers 1 views
1.8 g of glucose is dissolved in 100 g of water in a beaker. At what temperature will the solution boil if the pressure is 1.013 bar ? Given that the

Correct Answer - 373.155 K `M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A)) or DeltaT_(b)=(K_(b)xxW_(b))/(M_(B)xxW_(A))` `W_(B)=1.8g, W_(A)=100g=0.1" Kg", M_(B)=180" g mol"^(-1), K_(b)=0.052" K kg mol"^(-1)` `DeltaT_(b)=((0.052" K kg mol"^(-1))xx(1.8 g))/((180" g mol"^(-1))xx(0.1" kg"))=0.005 K` Boiling point of pure...

2 Answers 1 views
300 gm, `30%`(w/w) NaOH solution is mixed with 500 gm `40%`(w/w) NaOH solution. What is `%` (w/v) NaOH if density of final solution is 2 gm/mL?

Correct Answer - A % (w/v) volume of NaOH`=((300xx0.3+500xx0.4)/(800))/2xx100=72.5`

2 Answers 1 views
Assertion(A) : Heat of neutralisation of `HCl` and `NaOH` is same as that of `H_(2)SO_(4)` with `NaOH`. Reason(R ) : `HCl`, `H_(2)SO_(4)` and `NaOH` a

Correct Answer - A Heat of neutralisation of strong acid with strong base is contant `(-137 Kcal//mol^(-1))`

2 Answers 1 views
Which of the following solutions have different `pH`? `100mL" of "0.2m" "HCl +100mL" of "0.4 M" "NH_(3)` `50mL" of "0.1 M" HCl +"50mL" of "03.2 M NH(3

(a) `{:(NH_(4)OH,+,HCl,rarr,NH_(4)Cl,+,H_(2)O),(0.2M,,0.1,,0.1,,0),(0.1,,0,,0.1,,0.1):}` `pOH=pK_(b)` (b) `{:(NH_(4)OH,+,HCl,rarr,NH_(4)Cl,+,H_(2)O),(0.1M,,0.05,,0,,0),(0.05,,0,,0.05,,):}` `pOH=pK_(b)+log(((0.05)/(0.05)))=pK_(b)` c `{:(NH_(4)OH,+,HCl,rarr,NH_(4)Cl,+,H_(2)O),(0.3,,0.15,,,,),(0.15,,0,,0.15,,):}` `pOH=pK_(b)` so all solution have same `pH`.

2 Answers 1 views
Calculate the work done when 50 g of iron is dissolved in HCl at `25^(@)C` in (i) a cloed vessel and (ii) an open beaker when the atmospheric pressure

(i) When the reactiion is carried in a closed vessel, the change in volume is zero. Hence, the work done by the system will be zero. (ii) when iron dissolves...

2 Answers 1 views
When 100 mL each of HCl and NaOH solutions are mixed, 5.71 kJ of heat was evolved. What is the molarity of two solution? The heat of neutralisation of

Correct Answer - 1M

2 Answers 1 views
`150 mL` of `0.5N HCl` solution at `25^(@)C` was mixed with `150 mL` of `0.5 N NaOH` solution at same temperature. Calculate the heat of neutralizatio

Correct Answer - `-16 kcal`.

2 Answers 5 views
If 2M, 200ml HCl, 2M, 100ml `CaCl_(2)` and 5M, 200ml `AlCl_(3)` is mixed then final concentration of `Cl^(-)` will be:

`[Cl^(-)]=(M_(1)V_(1)+2M_(2)V_(2)+3M_(3)V_(3))/(V_(1)+V_(2)+V_(3))` `(2xx200+2xx2xx100+3xx5xx200)/(200+100+200)` `=(400+400+3000)/(500)=(3800)/(500)=7.6M`

2 Answers 1 views
Prove that `^100 C_2^(100)C_2+^(100)C_2^(100)C_4+^(100)C_4^(100)C_6++^(100)C_(98)^(100)C_(100)=1/2[^(200)C_(98)-^(100)C_(49)]dot`

To find `S = .^(100)C_(0).^(100)C_(2)+.^(100)C_(2).^(100)C_(4)+.^(100)C_(4).^(100)C_(6)+"....."+.^(100)C_(98).^(100)C_(100)` Consider, `.^(100)C_(0).^(100)C_(2)+.^(100)C_(1).^(100)C_(3) + .^(100)C_(2).^(100)C_(4)+.^(100)C_(3)+.^(100)C_(5)+"...." + .^(100)C_(98).^(100)C_(100)` `= .^(100)C_(0).^(100)C_(98)+.^(100)C_(1).^(100)C_(97) + .^(100)C_(2).^(100)C_(96)+.^(100)C_(3).^(100)C_(95) + "....."+^(100)C_(98) .^(100)C_(0)` = Coefficients of `x^(98)` in `(1+x)^(100) (1+x)^(100)` `= .^(200)C_(98) " "(1)` Also,...

2 Answers 1 views
X