A metal `M` which is not affected by strong acids like conc. `HNO_(3),conc.H_(2)SO_(4)` and conc. Solution of alkalies like `NaOH,KOH` forms `MCI_(3)`

Monjur Alahi

A metal `M` which is not affected by strong acids like conc. `HNO_(3),conc.H_(2)SO_(4)` and conc. Solution of alkalies like `NaOH,KOH` forms `MCI_(3)` which finds use for toning in photography. The metal `M` is
A. Ag
B. Hg
C. Au
D. Cu


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Salim Ahmed Kawsar

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Correct Answer - C
`Au to` Noble metal `&` it is used in tonning for photography.

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The heat of neutrasation of a strong acid and a strong alkali is `57.0KJ mol^(-1)` . The heat released when `0.5` mole of `HNO_(3)` solution is mixed

Correct Answer - B Here, `0.2` moles of `NaOH` is limiting reagent. Therefore, amount of heat liberated by neutralisation of `0.20` moles of `NaOH` is `=5.70xx0.2=11.4kJ`.

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Aniline on heating with conc. `HNO_(3)+` conc. `H_(2)SO_(4)` mixture yields

Correct Answer - C Since `HNO_(3)` is also a strong oxidising agent, therefore, aniline forms tarry oxidation products along with some oxidation products.

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In the reaction of benzene with a mixture of conc. `HNO_(3)` and conc. `H_(2)SO_(4)`, the active species involved is

Correct Answer - C In nitration of benzene, the reactive species in - volved is nitronium ion `(NO_(2)^(+))`.

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Nitrobenzene can be prepared from benzene by using a mixture of conc. `HNO_(3)` and conc. `H_(2)SO_(4)`. In the nitrating mixture, nitric acid acts as

Correct Answer - A `HNO_(3)+H_(2)SO_(4) rarr NO_(2)^(+)+HSO_(4)^(-)+H_(2)O` In this reaction nitric acid acts like a base.

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What volume of aqueous solution of NaOH that is `80%` by mass NaOH, contanis 0.4 mol of NaOH. Density of solution is `0.8 gm//ml`.

Correct Answer - B `0.4 " mole " NaOH = 0.4 xx 40 = 16 gm` `80%` by mass NaOH `80 gm NaOH` in 100 gm solution 16 gm NaOH in...

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Cancerntrated `HNO_(3)` used in the laboratory is usually 69% by mass of `HNO_(3)`. The density of `HNO_(3)`. The density of `HNO_(3)` solution is 1`.

Correct Answer - 15.44 M `"Mass of solution"= 100 g, "Mass os "HNO_(3)=69 g` `"Volume of solution"=(100 g)//(1.41" g cm"^(-3))=70.92 cm^(3)=0.07092 dm^(3)` `"Molar mass of "HNO_(3)=1 +14 +3xx16=63" g mol"^(-1)` `"Molarity...

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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`

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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))`

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Calcutate `pH` of following solutions: (i) 0.001 M HNO_(3),(ii) 0.005m `H_(2)SO_(4)` ,(iii) 0.01 M KOH (iv) `10^(-6)` M NaOH, (v)0.0008 M ` Ba(OH)_(2)

`(i) {:(0.001MHNO_(3),:,HNO_(3)rarr,H^(+),+,NO_(3)^(-)),(,t=0,10^(-3)M,0,,10^(-3)),(t=eq,,0,10^(-3),,10^(-3)):}` ` :. pH=-log10^(-3)=3.` `(ii){:(0.001MH_(2)SO_(4),:,H_(2)SO_(4)rarr,SO_(4)^(2-),+,2H^(+)),(,t=0,0.005M,0,,0),(t=eq,,0,0.005,,10^(-2)):}` ` :.pH=log(10^(-2))=2.` `(iii){:(0.001MKOH,:,KOHrarr,K^(+),+,OH^(-)),(,t=0,10^(-2)M,0,,0),(t=eq,,0,10^(-2),,2xx10^(-2)):}` iv`{:(10^-8MNaOH,:,NaOHrarr,Na^(+),+,OH^(-)),(,t=0,10^(-8)M,0,,0),(t=eq,,0,10^(-8),,10^(-8)),(t=eq,,H_(2)Orarr,H^(+),OH^(-),),(,,,x,,x):}` So, in solution `[OH^(-)]=(10^(-8)+x)M` and `[H^(+)]=x` `rArr=x=9.5xx10^(-8)=[H^(+)] :. pH=7.022` `(V){:(0.0008MBa(OH)_(2),:,Ba(OH)_(2)rarr,Ba^(2+),+,2OH^(-)),(,t=0,8xx10^(-4)M,0,,0),(t=eq,,0,8xx10^(-4),,2xx8xx10^(-4)):}` `[OH^(-)]=16xx10^(-4)M` `:. pOH=2.8` `:. pH=11.2`

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Upon titrating `50mL` of `1.96%(w//v) H_(2)SO_(4)` with a `KOH` solution (containing `11.2g KOH` per litre of solution on adding `50mL KOH` solution.

`[H_(2)SO_(4)]=(19.6)/(98)=0.2 M` and `KOH=(11.2)/(56)=0.2M` so for equivalent point, `(0.2xx50xx2)/(0.2)=100mL` of `KOH`solution should be added. `[H^(+)]=(0.2xx2-0.2)/(2)=0.1M` so `pH=1`.

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