One mole of an organic amide (A) on alkaline hydrolysis gives one mole of `NH_(3)` and one mole of a monobasic acid of equivalent mass 74. What is the molecular mass of (A) ?
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`Ag^(+) + NH_(3) ltimplies [Ag(NH_(3))]^(+), k_(1)=6.8 xx 10^(-5)` `[Ag(NH_(3))]^(+) + NH_(3) ltimplies [Ag(NH_(3))_(2)]^(+)`, `k_(2) = 1.6xx10^(-3)`
Correct Answer - a k = `k_1 xx K_2` = `(6.8 xx 10^(-5)) xx (1.6 xx 10^(-3))` = `1.08 xx 10^(-7)`.
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The enthalpies of neutralization of a weak acid HA & a weak acid HB by `NaOH` are -6900 Cal/equivalent & -2900 Cal/equivalent repsectively. When one e
Correct Answer - `1 : 3` 1 gram eq. of `NaOH` reacts with `HA` and `HB` a gram eq. of `NaOH` reacts with `HA` b gram eq. of `NaOH` react with...
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1 g of a monobasic acid dissolved in 200 g of water lowers the freezing point by `0.186^@ C`.On the other hand when 1 g of the same acid is dissolved
Correct Answer - 60 Let molar mass of acid be m. then 0.186=`ixx1/mxx1000/200xx1.86` Also, `1/m=125xx0.1xx10^(-3) so m=10000/125=400/5=80` so `i=80/50=8/5=1.6=1+ alpha implies alpha=60%`
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At `-50^(@)C` liquid `NH_(3)` has ionic product is `10^(-30)` .How many amide `(NH_(2)^(-))` ions are present per mm`.^(3)` in pure liqudi `NH_(3)`? (
`K=[NH_(4)""^(+)][NH_(2)""^(=)]=10^(-30)` `[NH_(2)^(-)]=[NH_(4)^(+)]=10^(-15)M " "( :.2NH_(3)hArr NH_(4)^(+)+NH_(2)""^(-))` No. of `NH_(2)^(-)` ions `NH_(2)^(-)=((10^(-15)mol e)/(L))((1L)/(10^(6)mm^(3)))((6xx10^(23)ions)/(mol e))=600ions//mm^(3)`
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The sodium salt of a certain weak monobasic organic acid is hydrolysed to an extent of 3% in its 0.1 M solution at `25^@C`.Given that the ionic produc
`h=0.3 C=0.1M` `K_(h)=CH^(2)=9xx10^(-5)` `K_(h)=(K_(w))/(K_(a))=9xx10^(-5)rArr K_(a)=(10^(-14))/(9xx10^(-5))=1.11xx10^(-10)~~1xx10^(-10)`
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What concentration of `OH^(-)` ions will reduce `NH_(4)^(+)` ion to `2xx10^(-5)M` in `0.4M` solution of `NH_(4)OH ? K(b)(NH_(4)OH)=1.8xx10^(-5)`
`{:(,NH_(4)OHhArr,NH_(4)^(+),+,OH^(+)),(t=eq,0.4(1-alpha),2xx10^(-5),,),(,,(=0.4alpha),,):}` So,`alpha=5xx10^(-5)(ltlt1):.0.4(1-alpha)~~0.4` `K_(b)=(2xx10^(-5)xx[OH^(-)])/(0.4)=1.8xx10^(-5)rArr [OH^(-)]=0.36M`
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Calculate the pH of `0.5L` of a`0.2 M NH_(4)Cl-0.2 M NH_(3)` buffer before and after addition of (a)` 0.05` mole of `NaOH` and (b) `0.05` mole of `HCl
`pOH=4.74+log((0.1)/(0.1))=4.74` `pH=9.26` (a) `{:(,NH_(4)^(+),+,H^(+),rarr,NH_(4)^(+)),(t=0,0.1"mole",,0.05"mole",,0.1"mole"),(,0.05"mole",,-,,0.15"mole"):}` `pOH=4.74+log((pOH=4.74+log((0.15)/(0.05))=4.26` `:. pH=9.74` (b)` {:(,NH_(3)^(+),+,H^(+),rarr,NH_(4)^(+)),(t=0,0.1"mole",,0.05"mole",,0.1"mole"),(,0.05"mole",,-,,0.15"mole"):}` `pOH=4.74+log((0.15)/(0.05))=5.22` `:.pH=8.78`.
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Percent yield of `NH_(3)` in the following reaction is 80% `NH_(2)CONH_(2) + 2NaOH overset(Delta)rarrNa_(2)CO_(3)+2NH_(3)` `6g NH_(2)CONH_(2)` reacts
Correct Answer - B `NH_(2)CONH_(2)+2NaOHoverset(80%)rarrNa_(2)CO_(3)+2NH_(3)` `0.1 " mol" " "0.2 " mol"` `(n_("urea"))/(1)xx(80)/(100)=n_(NH_(3))/(2)` `n_(NH_(3)) = 0.16` `n_(NH_(3)) = 0.16xx17 = 2.72g`
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A mixture of an acid anhydride (A) and a monobasic acid (B) on heating produces another monobasic acid (C) of equivalent weight 74 and an anhydride (D
Acid anhydride (A) +`underset((CF_(3) "group")) ("Monobasic acid (B)")` `hArr`Monobasic acid(C ) +`underset((Two CF_(3) "group")) ("Anhydride(D)")` Eq. wt.=128 Eq. wt.=74 Probable formula of acid (B)=`CF_(3)(CH_(2))_(n)COOH` Equivalent weight of (B) =`("Molecular weight")/("Basicity")`...
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An Organic compound (A) `C_(4)H_(7)Cl_(3)` yields (B) with KOH(aq.) (B) on treating with `NH_(3)` gives (C) which on heating gives (D). On heating (D)
(A) `(CH_(3))_(2)CH"CC"l_(3)`, (B) `(CH_(3))_(2)CHCOOH`, (C) `(CH_(3))_(2)CHCOONH_(4)`, (D) `(CH_(3))_(2) CHCONH_(2)`
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