341 + Interview Questions in STOICHIOMETRY IN GENERAL AWARENESS Page 4 InterviewSolution

151.	A `2.0 g` sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of `CO_(2)` ceases. The volume of `CO_(2)` at `750mm Hg` pressure and at `298 K` is measured to be `123.9 mL`. A `1.5 g` of the same sample requires `150 mL` of `(M//10) HCl` for complete neutralisation. Calculate the percentage composition of the components of the mixture.
Answer» Only `NaHCO_(3)` decoposes to give `CO_(2)`. `2NaHCO_(3)overset(Delta)toNa_(2)CO_(3)+CO_(2)+H_(2)O` `PV=nRT` or `(750)/(760)xx(123.9)/(1000)=nxx0.0821xx298` or `n=4.997xx10^(-3)approx5xx10^(-3) mol CO_(2)` Moles of `NaHCO_(3)=5xx10^(-3)xx2=0.01` Mass of `NaHCO_(3)=0.01xx84=0.84g` Percentage of `NaHCO_(3)=(0.84)/(2)xx10=42%` Millimoles of `HCl=150xx(1)/(10)=15` HCl reacts with `Na_(2)CO_(3)` and `NaHCO_(3)`. `1.5 g` mixture has `(0.01)/(2)xx1.5=0.0075` mol `NaHCO_(3)` `=7.5 m" mol of "NaHCO_(3)` `NaHCO_(3)+"HCl"toNaCl+CO_(2)+H_(2)O` Millimoles of HCl neutralised by `Na_(2)CO_(3)=15-7.5=7.5` `Na_(2)CO_(3)+2"HCl"to2NaCl+CO_(2)+H_(2)O` Millimoles of `Na_(2)CO_(3)=(7.5)/(2)=3.75` Mass of `Na_(2)CO_(3)=3.75` Mass of `Na_(2)CO_(3)=3.75xx106xx10^(-3)=0.3975g` Percentage of `Na_(2)CO_(3)=(0.3975)/(1.5)xx100=26.5%` Percentage of `Na_(2)SO_(4)=100-(42+26.5)=31.5%`

Discussion

152.	Sodium bicarbonate on heating decomposes to form sodium carbonate, `CO_(2)`, and water. If 0.2 mole of sodium bicarbonate is completely decomposed, how many mołe of sodium carbonate is formed?A. 0.1B. 0.2C. 0.05D. 0.025
Answer» Correct Answer - A The reaction is involved is: `2NaHCO_(3)(s) overset(Delta)to Na_(2)CO_(3)(s)+CO_(2)(g)+H_(2)O(l)`

Discussion

153.	The equivalent mass of dilvalent metal is W. The molecular mass of its chloride is:A. W+35.6B. W+72C. 2W+72D. 2W+35.6
Answer» Correct Answer - C

Discussion

154.	Calculate the weight of `MnO_(2)` and the volume of HCl of specific gravirty 1.2 g g `mL^(-1)` and `5%` by weight needed to produce 1.12 L of `Cl_(2)` at STP by the reaction `MnO_(2)+4"HCl"toMnCl_(2)+3H_(2)O+Cl_(2)`
Answer» (a). `N_(HCl)=(% by weightxx10xxd)/(Ew of HCl)` `=(5xx10xx1.2)/(36.5)=1.64` Now, m" Eq of "`MnO_(2)-=m" Eq of "HCl` `-=m" Eq of "Cl_(2) formed`. `-=(1.12)/(11.2)xx10^(3)` `=100[{:("Ew of "Cl_(2)=(M)/(2)),(1 " Eq of "Cl_(2)=11.2L):}]` (b). Volume of HCl used: `NxxV=100` `1.64xxV=100` `thereforeV_(HCl)to60.97mL` Bacause HCl is also used to give `MnCl_(2)` thus volume used is double that required for the reduction of `MnO_(2)`. `V_(HCl)=2xx60.97=121.94mL` (c). Also, m" Eq of "`MnO_(2)=m" Eq of "HCl=100` `(W)/((87)/(2))xx10^(3)=100(Ew of MnO_(2)=(55+32)/(2))` `thereforeW_(MnO_(2))=4.35g`

Discussion

155.	What volume of hydrogen will be liberated at NTP by the reaction of Zn on 50mL dilute `H_(2)SO_(4)` of specific gravity 1.3 and having purity 40%?A. 3.5litreB. 8.25litreC. 6.74litreD. 5.94litre
Answer» Correct Answer - D Mass of `H_(2)SO_(4) = (50xx1.3xx40)/(100) = 26 g` `zn + H_(2)SO_(4) to ZnSO_(4) + H_(2)` Volume of `H_(2)` at`NTP = (22.4)/(98) xx 26 = 5.94` litre]

Discussion

156.	16 g of `SO_(x)` gas occupies 5.6 L at 1 atm and 273 K.What will be the value of x ?
Answer» Correct Answer - 2

Discussion

157.	0.58 g of `CH_3(CH_2)_(c)COOH` was burnt in excess air and the resulting gases `(CO_2` and `H_2O`) were passed through excess NaOH solution. The resulting solution was divided into two equal parts. One part requires 50 " mL of " 1.0 M HCl for complete neutralisation using phenolphthalein indicator. Another part required 80 " mL of " same HCl for neutralisation using methyl orange as indicator. Calculate the value of n and the amount of excess NaOH solution taken initially.
Answer» `CH_(3)(CH_(2))_(n)COOH+[(3n+4)/(2)]O_(2)to(n+2)CO_(2)+[(2n+4)/(2)]H_(2)O` 1 " mol of "acid`=(n+2) " mol of "CO_(2)` `(0.58)/((60+14n)) " mol of "acid=((n+2)xx0.58)/((60+14n))` " mol of "`CO_(2)` This `CO_(2)` is passed in excess `NaOH` where `CO_(2)` is converted to `Na_(2)CO_(3)` and some NaoH is left With phenolphthalein: `(1)/(2) m" Eq of "Na_(2)CO_(3)+m" Eq of "NaOH` left`=m" Eq of "HCl` `=50xx1xx1` `=50` `i.e., (a)/(2)+b=50` ..(i) With methyl orange `a+b=80xx1xx1` ..(ii) By equations (i) and (ii) we get `b(m" Eq of "NaOH` left)`=20` `a(m" Eq of "Na_(2)CO_(3) formed)=80-20=60` Total m" Eq of "`NaOH used=20+60=80` Weight of NaOH (total)`=80xx10^(-3)xx40=3.2g` Also `m" Eq of "Na_(2)CO_(3)=m" Eq of "CO_(2)` `60=((n+2)xx0.58)/((60+14n))xx2`(n-factor)`xx10^(3)` `thereforen=4`

Discussion

158.	A solution contaning `2.7 xx 10^(-3) mol of A^(2+) "ion required" 1.6 xx 10^(-3)mol of MnO_(4)^(2-)` for the oxidation of `A^(2+) "to" AO_(3)^(-)` the medium is:A. neutralB. acidicC. strong basicD. none of these
Answer» Correct Answer - B

Discussion

159.	What will be the normally of a solution obtained by mixiing 0.45N and 0.60N NaOH in the ration 2:1 by volume?A. 0.4NB. 0.5NC. 1.05ND. 0.15N
Answer» Correct Answer - B

Discussion

160.	The ration of mass per cent of C and H of an organic compound `(C_(x)H_(y)O_(z)) "is"6:1`. If one molecule of the above compound `(C_(x)H_(Y)O_(z))` contains half as much oxygen as required to burn one molecule of compound `C_(x)H_(Y)` compleltely to `CO_(2) and H_(2)O`. The empirial formula of compound `C_(x)H_(y)O_(z)` is:A. `C_(2)H_(4)O`B. `C_(3)H_(4)O_(2)`C. `C_(2)H_(3)O_(3)`D. `C_(3)H_(6)O_(3)`
Answer» Correct Answer - C Ratio of mass prcentage of carbon and hydrogen in the gives organic compound = 6:1 `therefore" " (12x)/(y)=(6)/(1)" " i.e, " " 2x = y` Number of oxygen atom in the compound`C_(x)H_(y)O_(z) = z " "......(i)` Equation of combustion of the compound `C_(x)H_(y) ` is : `C_(x)H_(y) + (x +(x)/(y)) O_(2) to xCO_(2)(g) +(y)/(2) H_(2)O(l)` Number of oxgyen atoms required for conbustion of `C_(x)H_(y) = (2x+(y)/(2))" ".....(i)` From (i) and (ii) ` z = (1)/(2) (2x +(y)/(2))= (1)/(2) (2x + (2x)/(2)) = (3x)/(2)` ` therefore " " x : y: z` ` X: 22x ":(3x)/(2)`

Discussion

161.	1 gram of carbonate `(M_(2)CO_(3))` on treatment with excess HCl produces 0.1186 mole of `CO_(2)`. The molar mass of `M_(2)CO_(3) "in g mol"^(-1)`A. 1186B. 84.3C. 118.6D. 11.86
Answer» Correct Answer - B The reaction is : `M_(2)CO_(3)+ 2HCl to 2Mcl + H_(2)O + CO_(2)` Number of moles of `CO_(2)` = Number of moles of `M_(2)CO_(3)` `0.01186 = ("Mass")/("Molar mass of" M_(2)CO_(3))= (1)/("Molar mass of"M_(2)CO_(3))` `therefore` Molar mass of `M_(2)CO_(3) = (1)/(0.01186) = 84.3 g mol^(-1)`

Discussion

162.	A mixture of FeO and `Fe_(2)O_(3)` is completely reacted with 100 mL of 0.25 M acidified `KMnO_(4)` solution. The resultant solution was then treated with Zn dust which converted `Fe^(3+)` of the solution to `Fe^(2+)`.The `Fe^(2+)` required 1000 mL of 0.10 `MK_(2)Cr_(2)O_(7)` solution. Find out the weight % `Fe_(2)O_(3)` in the mixture.A. 80.85B. 19.15C. 50D. 89.41
Answer» Correct Answer - A

Discussion

163.	20 " mL of " `H_(2)O_(2)` is reacted completely with acidified `K_(2)Cr_(2)O_(7)` solution 40 " mL of " `K_(2)Cr_(3)O_(7)` solution was required to oxidised the `H_(2)O_(2)` completely. Also, 2.0 " mL of " the same `K_(2)Cr_(2)O_(7)` solution required 5.0 " mL of " a 1.0 M `H_(2)C_(2)O_(4)` solution to reach equivalence point. Which of the following statements is/are correct?A. The `H_(2)O_(2)` solution is 5 M.B. The volume strength of `H_(2)O_(2)` is 56V.C. The volume strength of `H_(2)O_(2)` is 112V.D. If 40 " mL of " `(5M)/(8H_(2)O_(2))` is further added to the 10 " mL of " above `H_(2)O_(2)` solution the volume strength of the resulting solution is changed to 16.8 V.
Answer» Correct Answer - A::B::D (i). `mEq H_(2)O_(2)-=m" Eq of "Cr_(2)O_(7)^(2-)` `(n=2)(n=6)` `20mLxxN_(1)=40mL xxN_(2)` (a). `m" Eq of "Cr_(2)O_(7)^(2-)-=m" Eq of "C_(2)O_(4)^(2-)(n=2)` `2.0xxN_(2)-=5.0xx1.0xx2` `N_(2)(Cr_(2)O_(7)^(2-))=5` ..(ii) Therefore substituting the `N_(2)` in equation (i) `20 mLxxN_(1)=40mLxx5` `N_(1)(H_(2)O_(2))=10` `M_(1)(H_(2)O_(2))=(10)/(2)=5M` (b). `1 N H_(2)O_(2)=5.6V` `therefore10 N H_(2)O_(2)=56V` (d). `N_(1)V_(1)+N_(2)V_(2)=N_(3)V_(3)(V_(3)=10+40=50mL)` `10xx10+40xx(5)/(8)xx2=N_(3)xx50` `N_(3)` (final) `H_(2)O_(2)=3N` The volume strength of `H_(2)O_(2)=5.6xx3=16.8V`

Discussion

164.	When 100 " mL of " 0.1 M `Ba(OH)_(2)` is neutralised with a mixture of x " mL of " 0.1 M HCl and y " mL of " 0.2 M `H_(2)SO_(3)` using methyl orange indicator what is value of x and y?A. 200100B. 100200C. 300200D. 200300
Answer» Correct Answer - A Methyl orange indicator indicates complete ionisation of HCl but first step ioonisation of `(H_(2)SO_(3)toH^(o+)+HSO_(3)^(ɵ)(n=1)(N=Mxx1)` First case: `Ba(OH)_(2)-=HCl` `N_(1)V_(1)=N_(2)V_(2)` `0.1xx2xx100-=x xx0.1xx1` `x=200mL` Second case: `Ba(OH)_(2)-=H_(2)SO_(3)` `N_(1)V_(1)=N_(2)V_(2)` `(0.1xx2)xx100=0.2xx1xxy` `x=200mL` `y=100mL`

Discussion

165.	`1M HCl and 2 M HCl ` are mixed in volume ratio 4:1. What is the final molarity of HCl solution?A. 1.5B. 1C. 1.2D. 1.8
Answer» Correct Answer - C

Discussion

166.	Which of the following is/are correct about the redox reaction? `MnO_(4)^(ɵ)+S_(2)O_(3)^(2-)+H^(o+)toMn^(+2)+S_(4)O_(6)^(2-)`A. 1 " mol of "`S_(2)O_(3)^(2-)` is oxidised by 8 " mol of "`MnO_(4)^(ɵ)`B. The above redox reaction with the change of pH from 4 to 10 will have an effect on the stoichiometry of the reaction.C. Change of pH form 4 to 7 will change the nature of the product.D. At `pH=7,S_(2)O_(3)^(2-)` ions are oxidised to `HSO_(4)^(ɵ)`
Answer» Correct Answer - B::C::D (a). `5e^(-)+MnO_(4)^(ɵ)toMn^(2+)(n=5)` `2S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+2e^(-)(n=(2)/(2)=1)` " Eq of "`MnO_(4)^(ɵ)-=" Eq of "S_(2)O_(3)^(2-)` `5xx` moles of `MnO_(4)^(ɵ)-=1xx` moles of `S_(2)O_(3)^(2-)` `therefore` 1 " mol of "`S_(2)O_(3)^(2-)=5 " mol of "MnO_(4)^(ɵ)` (b). pH changes from 4 to 10 (acidic to strongly basic) `e^(-)+MnO_(4)^(ɵ)toMnO_(4)^(2-)(n=1)` `S_(2)O_(3)^(2-)to2SO_(4)^(2-)+8e^(-)(n=8)` " Eq of "`MnO_(4)^(ɵ)=" Eq of "S_(2)O_(3)^(2-)` `therefore` 1 " mol of "`S_(2)O_(3)^(2-)=(1)/(8)" mol of "MnO_(4)^(ɵ)` Hence with change of pH from 4 to 10, will change the stoichiometry of reaction and also changes the product. (c). pH changes from 4 to 7 (acidic to neutral medium) `3e^(-)+MnO_(4)^(ɵ)toMnO_(2)(n=3)` `S_(2)O_(3)^(2-)to2HSO_(4)^(ɵ)+8e^(-)(n=8)` Hence it will also effect the stoichiometry of reaction and natural of product. (d). `At pH=7,S_(2)O_(3)^(2-)` is oxidised to `HSO_(4)^(ɵ)` ion.

Discussion

167.	100 " mL of " 0.2 M Kal`(OH)_(2)CO_(3)` solution is completely neutralised by a standard solution of `(M)/(4)H_(2)C_(2)O_(4)`. Which of the following is/are wrong?A. The volume of `H_(2)C_(2)O_(4)` required is 160 mL.B. the volume of `H_(2)C_(2)O_(4)` required 80 mL.C. The normality of `KAl` `(OH)_(2)CO_(3)` is 0.4 ND. It is a redox reaction.
Answer» Correct Answer - B::C::D n factor for `overset(+1+3)(Kal)(overset(-2)(OH))_(2)overset(-2)(CO_(3))` is the total positive or negative charge`=4`. n factor for `H_(2)C_(2)O_(4)=2(2H^(o+)` ions) It is an acid-base titration `m" Eq of "Kal(OH)_(2)CO_(3)=m" Eq of "H_(2)C_(2)O_(4)` `100mL xx0.2xx4=(1)/(4)xx2xxV mL` `V_(H_(2)C_(2)O_(4))=160 mL` `N_(KAl(OH)_(2)CO_(3))=0.1xx4=0.8N`

Discussion

168.	What volume of 0.10 M `H_(2)SO_(4)` must be added to 50 mL of a 0.10 NaOH solution to make a solution in which molarity of the `H_(2)SO_(4)` is 0.050M?A. 400mLB. 200mLC. 100mLD. none of these
Answer» Correct Answer - C

Discussion

169.	The vapour density of mixture consisting of `NO_2` and `N_2O_4` is 38.3 at `26.7^@C`. Calculate the number of moles of `NO_2` I `100g` of the mixture.
Answer» Molecular weight of the mixture`=38.3xx2=76.6` Molecular weight of `NO_2=46 and N_2O_4=92` Let the number of moles of `NO_2` be x. Number of moles of `N_2O_4=(100-46x)/(92)` `x+(100-46x)/(92)=(100)/(76.6)impliesx=0.438mol`

Discussion

170.	The oxidation number of sulphur in `S_(8),S_(2)F_(2),H_(2)S and H_(2)SO_(4)` respectively are:A. `0, +1, -2 and 6`B. `+2,0, +2 and 6`C. `0, +1, +2 and 6`D. ` -2,0, +2 and 6`
Answer» Correct Answer - A

Discussion

171.	`RH_(2)` ( ion exchange resin) can replace `Ca^(2+)`d in hard water as. `RH_(2)+Ca^(2+)rarrRCa+2H^(+)` `1 "litre"` of hard water passing through `RH_(2)` has `pH2`. Hence hardness in `pp m "of" Ca^(2+)` is:A. 200B. 100C. 50D. 125
Answer» Correct Answer - A `pH=2,[H^(o+)]=10^(-2)M` `[Ca^(2+)]=(10^(-2))/(2)=0.5xx10^(-2)xx40gL^(-1)` `=(0.5xx10^(-2)xx40xx10^(6))/(10^(3))(g)/(10^(6))mL=200ppm`

Discussion

172.	One commercial system removes `SO_(2)` emission from smoke at `95(@)C` by the following set of reaction : `SO_(2)(g)+Cl_(2)(g) to SO_(2)Cl_(2)`(g) `SO_(2)Cl_(2)(g) +H_(2)O(l) to H_(2)SO_(4)+HCl` `H_(2)SO_(4)+Ca(OH)_(2) to CaSO_(4)+H_(2)O` How many grams of `CaSO_(4)` may be produced from 3.78 g of `SO_(2)` ?
Answer» Correct Answer - 8

Discussion

173.	In which fo the following has the oxidation number of oxygen been arragned in increasing order ?A. `OF_(2) lt KO_(2) lt BaO_(2) lt O_(3)`B. `BaO_(2) lt KO_(2) lt O_(3) lt OF_(2)`C. `BaO_(2) lt KO_(2) lt OF_(2) lt KO_(2)`D. ` KO_(2) lt OF_(2) lt O_(3) lt BaO_(2)`
Answer» Correct Answer - B

Discussion

174.	1.67 g mixture of Al and Zn was completely dissolved in acid and evolved 1.69 L of `H_2` at STP. Calculate the weight Al and Zn in the mixture.
Answer» Let x and y be the weights of Al and Zn in the mixture. `x+y=1.67` .. (i) " Eq of "`Al+Eq` of `Zn= Eq` of `H_2` `((a)/((27)/(3))+(b)/((65)/(2)))=(1.69)/((22.4)/(2)) ` …(ii) Solving equation (i) and (ii) we get `x=1.25 g` `y=0.42`

Discussion

175.	If it is known that in `Fe_(0.96)O`, Fe is present in +2 and +3 oxidation state, what is the mole fraction of `Fe^(2+)` in the compound?A. `(12)/(25)`B. `(25)/(12)`C. `(1)/(12)`D. `(11)/(12)`
Answer» Correct Answer - D

Discussion

176.	In acidic medium dichromate oin osxidizes stannous ion as : `xSn^(2+)+yCr_(2)O_(7)^(2-)+zH^(+) to aSn^(4+)+bCr^(3+)cH_(2)O`A. "the value of x:y is " 1:3B. the value of x+y+z is 18C. a:b " is " 3:2D. the value of z-c is 7
Answer» Correct Answer - B::C::D

Discussion

177.	The reaction, `2C+O_(2)rarr2CO` is carried out by taking `24 g` of carbon and `96 gO_(2)`, find out: (a) which reactant is left in excess? (b) How much of it is left? (c ) How many mole of `CO` are formed? (d) How many `g` of other reactant should be taken so that nothing is left at the end of reaction?
Answer» `underset(=24g)underset("2mol")(2Mg)2C(s)+underset(32g)underset(1"mol")(O_(2)) to underset(56g)underset(2"mol")(2CO(g))` Let carbon be completely consumed. 24g carbon give 56g CO. Let `O_(2)` give 56g of CO. `therefore 96"g"O_(2) "will give" (56)/(32)xx96g CO=168"g CO"` Since, carbon gives least amound of product, i.e. `56g of CO or 2` of mole CO, hence carbon will be the limiting reactant. Excess reactant is `O_(2)` Amound of `O_(2)` used=56-24=32g Amount of `O_(2)` left=96-32=64g 32g of `O_(2)` react with 24g carbon `therefore 9g of O_(2) ` will react with 72g carbon Thus, carbon should be taken 72g so that nothing is left at the end of the reaction.

Discussion

178.	100 g sample of calcicum carbonate is reaction with 70g of orthophosphoric acid. Calculate (a) the number of grams of calcium phosphate that could be produced. (b) the number of grams of excess reagent that will remain unreacted.
Answer» The balanced equation is `underset("=300g")underset(3(40+12+48))underset("3mol")(3CaCO_(3))+underset(=196g)underset(2(3+3+64))underset("2mol")(2H_(3)PO_(3)) to underset(=310g)underset((3xx40+2xx64))underset("1 mol")(Ca_(3)(PO_(4))_(2))+3CO_(2)+3H_(2)O` `300"g of"CaCO_(3) "produce" Ca_(3)(PO_(4))_(2)=310"g or 1 mol"` `100"g of"CaCO_(3) "would produce"` `Ca_(3)(PO_(4))_(2)=(310)/(300)xx100` =103g =0.03mol `196"g of "H_(3)PO_(4) "produce" Ca(PO_(4))_(2)=310"g of 1 mol"` `70"g of "H_(3)PO_(4) "produce" Ca(PO_(4))_(2)=(310)/(196)xx70` `=110.7g or 0.356"mol"` The above values suggest that `CaCO_(3)` is the limiting reagent. Hence, calcium phosphate formed is 103g of 0.33 mole. (b) For producing 103 g of `Ca_(3)(PO_(4))_(2), H_(3)PO_(4)` required will be `=(196)/(310)xx1033=65.12g` Mass of remaining `H_(3)PO_(4)=(70-65.12)=4.88g`

Discussion

179.	If 20g of `CaCO_(3)` is treated with 20g of HCl. How many grams of `CO_(2)` can be generated according to the following equations? `CaCO_(3)+2HCl(aq) to CaCl_(2)(aq.)+H_(2)O(l)+CO_(2)g`
Answer» `underset(100g)underset("1 mol")(CaCO_(3))+underset(73g)underset("2 mol")(2HCl(aq)) to CaCl_(2)(aq.)+H_(2)O(l)+underset(44g)underset("1 mol")(CO_(2)g)` Let `CaCO_(3)(s)` be completely consumed in the reaction. `therefore 100g CaCO_(3) "give" 44gCO_(2)` `therefore 20g CaCO_(3)"will give" (44)/(100)xx20g CO_(2)=8.8CO_(2)` Let HCl be completely consumed. `therefore 73g "HCl give" 44g CO_(2)` `therefore 20 "g of HCl will give" (44)/(73)xx20 g CO_(2)=12.054g CO_(2)` Since, `CaCO_(3)` gives least amount of product `CO_(2)`, hence, `CaCO_(3)` is limiting reactant. Amount of `CO_(2)` formed will be 8.8g

Discussion

180.	Calculate the percentage concentration of the remaining solution when 500 g of a `20%` solution by weight is cooled, 40 g of solute is precipitated.
Answer» Weight of solution `=500g` Weight of solute`=(20xx500)/(100)=100g` Weight of solute precipitated`=40g` Weight of solute left in the solution`=100-40=60g` Weight of solution left`=500-40=460g`

Discussion

181.	One mole of calcium phosphide on reaction with excess of water give:A. three moles of phosphineB. one mole of phosphrinc acidC. two moles of phosphineD. one mole of `P_(2)O_(5)`
Answer» Correct Answer - C `Ca_(3)P_(2)+6H_(2)O to 3Ca(OH)_(2)+2PH_(3)` `"1 mole of "Ca_(4)P_(2)-=2"moles of "PH_(3)`

Discussion

182.	Match the following
Answer» Correct Answer - A (A) `underset("10 g")underset("100 g") (CaCO_(3))overset(Delta)to CaCO + underset("0.224L")underset("22.4 L")(CO_(2))` (B) `underset("1.06g")underset("1 mol (106g)")(Na_(2)CO_(3))+ 2HCl to 2NaCl + H_(2)O+ underset("0.224 L")underset("22.4L")(CO_(2))` (C) `underset("12g")underset("2.4g")(C) + O_(2) to underset("0.448L")underset("22.4L")(CO_(2))` (D) `underset("0.56 g")underset("28 g")(CO) + (1)/(2)O_(2) to underset("0.448L")underset("22.4 L")(CO_(2))`

Discussion

183.	What volume of air containing 21% oxygen by volume is required to completely burn 1kg of carbon containing 100% combustible substances?
Answer» Combustion of carbon may be given as, `underset(12g)underset(1"mol")(C(s))+underset(1"mol")underset(1"mol")(O_(2)(g)) to CO_(2)(g)` `therefore "12 g of carbon required 1 mole" O_(2) "for complete combustion"` `therefore "1000g of carbon will requires" (1)/(12)xx1000 "mole" O_(2)"for combustion i.e. 83.33 mole" O_(2)` Volume of `O_(2)` at NTP=`83.33xx22.4"litre"=18.66.592"litre"` `therefore21 "litre" O_(2)` is present in 100 litre air `therefore 18.66.592 "litre" O_(2)` will be present in `(100)/(21)xx1866.592 "litre" O_(2) =8888.5"litre=8.8885xx10^(5)xx10^(3)"litre"`

Discussion

184.	How many grams of oxygen are required to burn completely 570g of octane?
Answer» `underset(2xx114)underset(2"mol")(2C_(8)H_(18))+underset(25xx32)underset("25 mole")(25O_(2))to16CO_(2)+18H_(2)O` First method: For burning `2xx114g` of octane, oxygen required `=25xx32g` Thus, for burning570 g of octane, oxygen required `(25xx32)/(2xx114)xx570g=2000g` Molar method: Number of moles of octane in 570grams `=(570)/(114)=5.0` For burning 2.0 moles of octane, oxygen required `=25"mol"=25xx32g` For buring 5 moles of octane, oxygen required `=(25xx32)/(2.0)xx5.0g=2000g` Proportion method: Let x g of oxygen be required for burning 570.0g of octane. It is known that `2xx114g` of teh octane require `25xx32g` of oxygen, then, the proportion, `(25xx32"g oxygen")/(2xx114"g octane")=(x)/("570.0g octane")` `therefore x=(25xx32xx570)/(2xx114)=2000g`

Discussion

185.	Chloringe is prepared in the laboratory by treating magnesse dixoide `(MnO_(2))` with aqueous hydrochlorine acid according to the reaction. `MnO_(2)+4HCl to MnCl_(2)+Cl_(2)+2H_(2)O`
Answer» 1 mole of `MnO_(2)` reacts with 4 mole of HCl or 87g `MnO_(2)` reacts with 146g HCl `therefore 5g MnO_(2) "will react with" (146)/(87)xx5"g HCl"=8.39g HCl`

Discussion

186.	(a). If both (A) and (R) are correct and (R) is the appropriate explanation for (A). (b). If both (A) and (R) are correct and (R) is the correct explanation of (A). (c). If (A) is correct but (R) is incorrect. (d). If (A) is incorrect, but (R) is correct. (e). If both (A) and (R) are incorrect. Q. Assertion (A): in the titration of `Na_(2)CO_(3)` with HCl using methyl orange indicator, the volume required at the equivalence point is twice that of the acid requried using phenolphthalein indicator. Reason (R): Two moles of HCl is requried for the complete neutralisation of 1 " mol of "`Na_(2)O_(3)`
Answer» Correct Answer - B `Na_(2)CO_(3)+"HCl"toNaHCO_(3)+NaCl` At this point phenolphthalein changes colour but when methyl orange is used as an indicator it changes colour when complete reaction takes place. `Na_(2)CO_(3)+2"HCl"to2NaCl+CO_(2)+H_(2)O`

Discussion

187.	Which of the following is/are incorrect for 17 g/L of `H_2O_2` solution ?A. Volume strengths is 5.6 at 273 K and 1 atmB. Molarity of solution is 0.5 MC. 1 mL of thios solution gives 2.8 mL `O_(2)` at 273 K and 2 atmD. The normality of solution is 2 N
Answer» Correct Answer - A::B::C

Discussion

188.	30 " mL of " a solution containing `9.15gL^(-1)` of an oxalate `K_(x)H_(y)(C_2O_4)_2.nH_2O` required for titration 27 " mL of " 0.12 N NaOH and 36 " mL of " 0.12 N `KMnO_4` for oxidation Find x,y,z. and n.
Answer» Normality of oxalate as an acid`=(27xx0.12)/(30)=0.108` Ew of oxalate as acid `=("strength")/(N)=(9.15)/(0.108)=84.72` Normality of oxalate as reducing agent`=(36xx0.12)/(30)` `=0.144` Ew of oxalate as reducing agent`=(9.15)/(0.144)=63.54` Oxalate as acid: `(Mw)/(y)=84.72` Oxalate as reducing agent `(Mw)/(2z)=63.54` `x+y=2z` `y=15z` `thereforex:y:z=1:3:2` `therefore39xx1+3xx1+2xx88+18n=254` `n=2` Formula is `KH_3(C_2O_4).2H_2O`

Discussion

189.	A mixture of `HCOOH` and `H_(2)C_(2)O_(4)` is heated with conc. `H_(2)SO_(4)`. The gas produced is collected and on treating with `KOH` solution the volume of the gas decreases by `1//6th`. Calculate molar ratio of two acids in original mixure.
Answer» `underset(1"mol")(HCOOH) underset("Heat")overset("Conc."H_(2)SO_(4))to underset(1"mol")(CO)+H_(2)O` `underset(1"mol")(H_(2)C_(2)O_(4)) underset("Heat")overset("Conc."H_(2)SO_(4))to underset(1"mol")(CO)+underset(1"mol")(CO_(2))+H_(2)O` Let a moles of HCOOH and b moles of `H_(2)C_(2)O_(4)` be present in the original mixture. `"Moles of CO formed "=a+b` `"Moles of " CO_(2) " formed "=b` Total moles of gases=a+b+b=a+2b `CO_(2)` is absorbed by KOH and the volume reduces by 1/6th. `"Moles of " =(a+2b)/(6)` `or b=(a+2b)/(6)` `or a=4b` `or (a)/(b)=4` `or a:b=4:1`

Discussion

190.	500 " mL of " 1.0 M `H_2C_2O_4`, 100 " mL of " 2.0 M `H_2SO_4`, and `40 g` of `NaOH` are mixed together, 30 " mL of " the above mixture is titrated against a standard solution of sodium carbonate containing 14.3 g of `Na_2CO_3.10H_2O` per 100 " mL of " solution. Find the volume of carbonate solution used for complete neutralisation.
Answer» m" Eq of "`H_2C_2O_4=2xx1.0xx500=1000` m" Eq of "`H_2SO_4=2xx2.0xx100=400` m" Eq of "acid(s) `=1400` Now m" Eq of "NaOH (base) `=(Wt.)/(Ew)xx100` `=(40)/((40)/(1))xx1000=1000` Excess of acid/30 mL in the mixture `=400` So m" Eq of "acid`=`m" Eq of "carbonate Normality of cabonate`=("strength")/(E)(143)/((286)/(2))` `((14.3)/(100mL)-=143gL^(-1))` `N=1` m" Eq of "acid (excess)/30 mL`=(30)/(600)xx400=20` `implies20=NV` `impliesV=20 mL` of carbonate

Discussion

191.	500 " mL of " a solution contains 2.65 g of `Na_2CO_3` and 4 g of `NaOH`. 20 " mL of " this solution titrated each time against `(N)/(10)H_2SO_4`. Find out the titre value if (a). Methyl orange is taken as an indicator (b). Phenolphthalein is taken as indicator.
Answer» `Na_2CO_4` in `1L=2xx2.65=5.3g` `NaOH` in `1L=2xx4=8g` Normality of `Na_2CO_3=(5.3)/(53)=0.1N` Normality of NaOH`=(8)/(40)=(1)/(5)N` `N_1V_1(H_2SO_4)=N_2V_2(Na_2CO_3)` `(N)/(10)xxV_1=20xx0.1N` `V_1=20cc` `N_1V_1(H_2SO_4)=N_2V_2(NaOH)` `(N)/(10)xxV_1=(N)/(5)xx20` `V_1=(1)/(5)xx20xx10=40cc` With methyl orange as an indicator: Full titre value of NaOH`+` Full titre value of `Na_2CO_3` `=40+20=60cc` With phenolphthalein as an indicator: Full titre value of `NaOH+(1)/(2)` titre value of `Na_2CO_3=40+10=50c c`

Discussion

192.	The neutralisation of a 1.20 g solution of a mixture of `H_2C_2O_4.2H_2O` and `KHC_2O_4.H_2O` and different impurities of a neutral salt consumed 37.80 " mL of " 0.25 N NaOH solution. On the other hand, on titration with `KMnO_4` for 0.40 g of the same substance, 43.10 " mL of " 0.125 N `KMnO_4` was required. Find the percentage composition of the substance being analysed.
Answer» Let the percentage of `H_2C_2O_4.2H_2O` be `a%` and the percentage of `KHC_2O_4.H_2O` be `b%` (a). `(1.2)/(100)((a)/(63)+(b)/(146))xx1000=37.80xx25` (b). `(0.40)/(100)((a)/(63)+(b)/(73))xx1000=43.10xx0.125` `a=H_2C_2O_4.2H_2O=14.3%` (b). `=KHC_2O_4.H_2O=81.7%`

Discussion

193.	0.5 g `CaBr_2` was dissolved in water and the solution is acidified with nitric acid, 50 " mL of " standard `0.1N` `AgNO_3` is added and the solution is shaken thoroughly, the remaining `Ag^(o+)` ions required 15 " mL of " 0.1 N `NH_4CNS` solution using ferric alum as the indicator. Calculate the percentage of `CaBr_2` in the sample.
Answer» `[Mw(CaBr_2)=40+2xx80=200g]` `[Ew(CaBr_2)=(200)/(2)=100g(n=2)]` `CaBr_2-=AgNO_3-=NH_4CNS` `mEq-=mEq-=mEq` `Br^(ɵ)-=Ag^(o+)-=CNS^(ɵ)` `mEq-=mEq-=mEq` m" Eq of "`CNS^(ɵ)=15xx0.1=1.5mEq` Excess of `Ag^(o+)` ions reacted with `CNS^(ɵ)-=1.5mEq` Total `Ag^(o+)=50xx0.1N=5mEq` `Ag^(o+)` ions reacted with `Br^(ɵ)=5-1.5=3.5mEq` `=3.5mEq` of `CaBr_2` Weight of `CaBr_2=3.5xx10^(-3)EqxxEw(CaBr_2)` `=3.5xx10^(-3)xx100=0.35g` `%` of `CaBr_2=(0.35xx100)/(0.5)=70%`

Discussion

194.	10 " mL of " 2 M HCl and 20 " mL of " 1 M `HNO_3` and V volume of `5M H_2SO_4` are mixed together and the solution was made upto 5 L. 10 " mL of " this acid solution exactly neutralises 28.6 " mL of " `Na_2CO_3` solution containing 1 g of `Na_2CO_4.10H_2O` in 100 " mL of " water. Calculate the amount of `SO_4^(2-)` ions in grams present in solution.
Answer» Mw of `Na_2CO_3.10H_2O=2xx23+12+3xx16+10xx18=286g` Ew of `Na_2CO_3.10H_2O("valency factor"=2)=(286)/(2)=143g` Strength of `Na_2CO_3.10H_2O=(1xx1000)/(100)=10gL^(-1)` `N_(Na_(2)CO_(3))=("Strength")/(Ew)=(10)/(143)` total mEq of mixture of acid in 5 L of solution`=10xx2xx1+20xx1xx1xxVxx5xx2` (n-factor) `=20+20+10V=40+10V` Now m" Eq of "acid in 10 mL solution `=m" Eq of "Na_2CO_3` used for it `=N_(Na_(2)CO_(3))xxV(mL)=(10)/(143)xx28.6` `=2mEq(i n 10mL)` `=(2xx5000)/(10)mEq(i n 5L)` `=1000mEq` m" Eq of "acid in 5 L solution`=1000mEq` `40+10V=1000` `10V=960` `10V-=960-=m" Eq of "H_2SO_4-=m" Eq of "SO_4^^(2-)` `m" Eq of "SO_4^(2-)=(weight)/(Ew)xx10^(3)=960` `(weightxx10^(3))/((96)/(2))=960` Weight of `SO_4^(2-)=(960xx48)/(1000)=4.6g`

Discussion

195.	100 " mL of " each thhree samples of `H_2O_2` labelled 2.8 vol 5.6vol, and 22.4 vol are mixed and then diluted with an equal volume of water. Calculate the volume strength of the resultant `H_2O_2` solution.
Answer» Volume strength of `H_2O_2=5.6xxN` `N_1 of 2.8 vol H_2O_2=(2.8)/(5.6)=0.5N` `N_2 of 5.6 vol H_2O_2=(5.6)/(5.6)=1N` `N_3 of 22.4 vol H_2O_2=(22.4)/(5.6)=4N` `N_1V_1+N_2V_2+N_3V_3=N_4V_4(V_4=300+300=600mL)` `0.5xx100+1xx100+4xx100=N_4xx600` `N_4=(50+100+400)/(600)=(550)/(600)=0.91N` Volume strength`=5.6xx0.91=5.09vol` Alternatively: (Volume strength`)_1+`(Volume strength`)_2` `+("Volume strengt"h)_3` `=(2.8+5.6+22.4)/(6)=(30.8)/(6)=5.1vol`

Discussion

196.	It requires `40.0mL` of `0.50M Ce^(4+)` to titrate `10.0 mL` of `1.0MSn^(2+) "to" Sn^(4+)`. What is the oxidation state of cerium in the reduced product?
Answer» `Ce^(4+)-=Sn^(2+)(Sn^(2+)toSn^(4+)+2e)(n=2)` `1mEq-=1mEq` `N_1V_1-=N_2V_2` `0.5xx`n-factor`xx40=1xx2`(n-factor)`xx10` `20n=20` `n=1` n-factor of `Ce^(4+)` is 1, i.e., it is reduced to `Ce^(3+)` `Ce^(4+)+e^(-)toCe^(3+)`

Discussion

197.	What is the normality of a solution that results from mixing 7.4 g of `Ca(OH)_2, 500` " mL of " `1 M HNO_3` and `10.0 mL` of `H_2SO_4` (specific gravity`=1.2,49%H_2SO_4` by weight)? The total volume of the solution was made to 1L after adding water?
Answer» (a). `Mw[Ca(OH)_2]=40+34=74` `7.4g. Of Ca (OH)_2=(7.4)/(74)-=0.1mol=10m` moles `=10xx2mEq` (n-factor`=2)` `-=20mEq` (b). `500 " mL of " 1 M HNO_3=500xx1xx1` (n-factor) `-=20mEq` (b). `500 " mL of " 1 M HNO_3=00xx1xx1` (n-factor) `-=500 mEq` (c). Normality of `H_2SO_4` `=(Wxx1000)/(Ewxx"Volume of solution in mL")` `{:[(Mw=98,"n-factor=2),(Ew=(98)/(2)=49g)]:}` or Normality of `H_2SO_4` `=(%"by weight"xx"density"("or specific gravity"))/(Ew)` `=(49xx10xx1.2)/(49)=12N` `m" Eq of "H_2SO_4=NxxV=12xx10=120 mEq` (d). Total m" Eq of "acid `=500+120=620 mEq` (e). m" Eq of "`Ca(OH_2)=20mEq` (f). m" Eq of "acid left `=620-20=600 mEq` (g). Normality of solution `=(mEq)/(V(mL))=(600)/(1000)=0.6N`

Discussion

198.	A bottle of `H_(2)O` is labelled as 10 vol `H_(2)O_(2)`. 112 " mL of " this solution of `H_(2)O_(2)` is titrated against 0.04 M acidified solution of `KMnO_(4)` the volume of `KMnO_(4)` in litre isA. 1 LB. 2 LC. 3 LD. 4 L
Answer» Correct Answer - A `5.6 V H_(2)O_(2)=1N` `10 V H_(2)O_(2)=(10)/(5.6)N` `m" Eq of "H_(2)O_(2)=m" Eq of "MnO_(4)^(ɵ)` `(10)/(5.6)xx112-=0.04xx5` ("n-factor")`xxV` `V-=1000ml=1L`

Discussion

199.	A 0.5 g sample containing `MnO_(2)` is treated with HCl liberating `Cl_(2)` is passed into a solution of KI and 30.0 " mL of " 0.1 M `Na_(2)S_(2)O_(3)` are required to titrate the liberated iodine. Calculate the percentage of `MnO_(2)` is the sample.
Answer» `MnO_(4)+4"HCl"toMnCl_(2)+Cl_(2)+2H_(2)O` `2KI+Cl_(2)to2KCl+I_(2)` `I_(2)+2Na_(2)S_(2)O_(3)toNa_(2)S_(4)O_(6)+2NaI` `m" mol of "Na_(2)S_(2)O_(3)=30xx0.1=3.0` `m" mol of "I_(2)=(3.0)/(2)=1.5` `m" mol of "`Cl_(2)=1.5` `m" mol of "MnO_(2)=1.5xx10^(-3)xx87=0.1305g` % of `MnO_(2)=(0.1305xx100)/(0.5)=26.1%`

Discussion

200.	One litre of a sample of hard water contain `4.44mg CaCl_(2) and 1.9mg "of" MgCl_(2)`. What is the total hardness in terms of ppm of `CaCO_(3)?`A. 2ppmB. 3ppmC. 4ppmD. 6ppm
Answer» Correct Answer - D

Discussion

Explore topic-wise InterviewSolutions in .

Sodium bicarbonate on heating decomposes to form sodium carbonate, `CO_(2)`, and water. If 0.2 mole of sodium bicarbonate is completely decomposed, how many mołe of sodium carbonate is formed?A. 0.1B. 0.2C. 0.05D. 0.025

The equivalent mass of dilvalent metal is W. The molecular mass of its chloride is:A. W+35.6B. W+72C. 2W+72D. 2W+35.6

Calculate the weight of `MnO_(2)` and the volume of HCl of specific gravirty 1.2 g g `mL^(-1)` and `5%` by weight needed to produce 1.12 L of `Cl_(2)` at STP by the reaction `MnO_(2)+4"HCl"toMnCl_(2)+3H_(2)O+Cl_(2)`

What volume of hydrogen will be liberated at NTP by the reaction of Zn on 50mL dilute `H_(2)SO_(4)` of specific gravity 1.3 and having purity 40%?A. 3.5litreB. 8.25litreC. 6.74litreD. 5.94litre

16 g of `SO_(x)` gas occupies 5.6 L at 1 atm and 273 K.What will be the value of x ?

A solution contaning `2.7 xx 10^(-3) mol of A^(2+) "ion required" 1.6 xx 10^(-3)mol of MnO_(4)^(2-)` for the oxidation of `A^(2+) "to" AO_(3)^(-)` the medium is:A. neutralB. acidicC. strong basicD. none of these

What will be the normally of a solution obtained by mixiing 0.45N and 0.60N NaOH in the ration 2:1 by volume?A. 0.4NB. 0.5NC. 1.05ND. 0.15N

1 gram of carbonate `(M_(2)CO_(3))` on treatment with excess HCl produces 0.1186 mole of `CO_(2)`. The molar mass of `M_(2)CO_(3) "in g mol"^(-1)`A. 1186B. 84.3C. 118.6D. 11.86

When 100 " mL of " 0.1 M `Ba(OH)_(2)` is neutralised with a mixture of x " mL of " 0.1 M HCl and y " mL of " 0.2 M `H_(2)SO_(3)` using methyl orange indicator what is value of x and y?A. 200100B. 100200C. 300200D. 200300

`1M HCl and 2 M HCl ` are mixed in volume ratio 4:1. What is the final molarity of HCl solution?A. 1.5B. 1C. 1.2D. 1.8

What volume of 0.10 M `H_(2)SO_(4)` must be added to 50 mL of a 0.10 NaOH solution to make a solution in which molarity of the `H_(2)SO_(4)` is 0.050M?A. 400mLB. 200mLC. 100mLD. none of these

The vapour density of mixture consisting of `NO_2` and `N_2O_4` is 38.3 at `26.7^@C`. Calculate the number of moles of `NO_2` I `100g` of the mixture.

The oxidation number of sulphur in `S_(8),S_(2)F_(2),H_(2)S and H_(2)SO_(4)` respectively are:A. `0, +1, -2 and 6`B. `+2,0, +2 and 6`C. `0, +1, +2 and 6`D. ` -2,0, +2 and 6`

`RH_(2)` ( ion exchange resin) can replace `Ca^(2+)`d in hard water as. `RH_(2)+Ca^(2+)rarrRCa+2H^(+)` `1 "litre"` of hard water passing through `RH_(2)` has `pH2`. Hence hardness in `pp m "of" Ca^(2+)` is:A. 200B. 100C. 50D. 125

In which fo the following has the oxidation number of oxygen been arragned in increasing order ?A. `OF_(2) lt KO_(2) lt BaO_(2) lt O_(3)`B. `BaO_(2) lt KO_(2) lt O_(3) lt OF_(2)`C. `BaO_(2) lt KO_(2) lt OF_(2) lt KO_(2)`D. ` KO_(2) lt OF_(2) lt O_(3) lt BaO_(2)`

1.67 g mixture of Al and Zn was completely dissolved in acid and evolved 1.69 L of `H_2` at STP. Calculate the weight Al and Zn in the mixture.

If it is known that in `Fe_(0.96)O`, Fe is present in +2 and +3 oxidation state, what is the mole fraction of `Fe^(2+)` in the compound?A. `(12)/(25)`B. `(25)/(12)`C. `(1)/(12)`D. `(11)/(12)`

In acidic medium dichromate oin osxidizes stannous ion as : `xSn^(2+)+yCr_(2)O_(7)^(2-)+zH^(+) to aSn^(4+)+bCr^(3+)cH_(2)O`A. "the value of x:y is " 1:3B. the value of x+y+z is 18C. a:b " is " 3:2D. the value of z-c is 7

100 g sample of calcicum carbonate is reaction with 70g of orthophosphoric acid. Calculate (a) the number of grams of calcium phosphate that could be produced. (b) the number of grams of excess reagent that will remain unreacted.

If 20g of `CaCO_(3)` is treated with 20g of HCl. How many grams of `CO_(2)` can be generated according to the following equations? `CaCO_(3)+2HCl(aq) to CaCl_(2)(aq.)+H_(2)O(l)+CO_(2)g`

Calculate the percentage concentration of the remaining solution when 500 g of a `20%` solution by weight is cooled, 40 g of solute is precipitated.

One mole of calcium phosphide on reaction with excess of water give:A. three moles of phosphineB. one mole of phosphrinc acidC. two moles of phosphineD. one mole of `P_(2)O_(5)`

Match the following

What volume of air containing 21% oxygen by volume is required to completely burn 1kg of carbon containing 100% combustible substances?

How many grams of oxygen are required to burn completely 570g of octane?

Chloringe is prepared in the laboratory by treating magnesse dixoide `(MnO_(2))` with aqueous hydrochlorine acid according to the reaction. `MnO_(2)+4HCl to MnCl_(2)+Cl_(2)+2H_(2)O`

Which of the following is/are incorrect for 17 g/L of `H_2O_2` solution ?A. Volume strengths is 5.6 at 273 K and 1 atmB. Molarity of solution is 0.5 MC. 1 mL of thios solution gives 2.8 mL `O_(2)` at 273 K and 2 atmD. The normality of solution is 2 N

30 " mL of " a solution containing `9.15gL^(-1)` of an oxalate `K_(x)H_(y)(C_2O_4)_2.nH_2O` required for titration 27 " mL of " 0.12 N NaOH and 36 " mL of " 0.12 N `KMnO_4` for oxidation Find x,y,z. and n.

A mixture of `HCOOH` and `H_(2)C_(2)O_(4)` is heated with conc. `H_(2)SO_(4)`. The gas produced is collected and on treating with `KOH` solution the volume of the gas decreases by `1//6th`. Calculate molar ratio of two acids in original mixure.

500 " mL of " a solution contains 2.65 g of `Na_2CO_3` and 4 g of `NaOH`. 20 " mL of " this solution titrated each time against `(N)/(10)H_2SO_4`. Find out the titre value if (a). Methyl orange is taken as an indicator (b). Phenolphthalein is taken as indicator.

100 " mL of " each thhree samples of `H_2O_2` labelled 2.8 vol 5.6vol, and 22.4 vol are mixed and then diluted with an equal volume of water. Calculate the volume strength of the resultant `H_2O_2` solution.

It requires `40.0mL` of `0.50M Ce^(4+)` to titrate `10.0 mL` of `1.0MSn^(2+) "to" Sn^(4+)`. What is the oxidation state of cerium in the reduced product?

What is the normality of a solution that results from mixing 7.4 g of `Ca(OH)_2, 500` " mL of " `1 M HNO_3` and `10.0 mL` of `H_2SO_4` (specific gravity`=1.2,49%H_2SO_4` by weight)? The total volume of the solution was made to 1L after adding water?

A bottle of `H_(2)O` is labelled as 10 vol `H_(2)O_(2)`. 112 " mL of " this solution of `H_(2)O_(2)` is titrated against 0.04 M acidified solution of `KMnO_(4)` the volume of `KMnO_(4)` in litre isA. 1 LB. 2 LC. 3 LD. 4 L

A 0.5 g sample containing `MnO_(2)` is treated with HCl liberating `Cl_(2)` is passed into a solution of KI and 30.0 " mL of " 0.1 M `Na_(2)S_(2)O_(3)` are required to titrate the liberated iodine. Calculate the percentage of `MnO_(2)` is the sample.

One litre of a sample of hard water contain `4.44mg CaCl_(2) and 1.9mg "of" MgCl_(2)`. What is the total hardness in terms of ppm of `CaCO_(3)?`A. 2ppmB. 3ppmC. 4ppmD. 6ppm