InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
A 1.2 g mixture of `Na_(2)CO_(3)` and `K_(2)CO_(3)` was dissolved in water to form `100 cm^(3)` of a solution. `20 cm^(3)` of this solution required `40 cm^(3)` of 0.1 N HCl for neutralisation. Calculate the mass of `Na_(2)CO_(3)` and `K_(2)CO_(3)` in the mixture. |
| Answer» Correct Answer - `Na_(2)CO_(3)=0.5962; K_(2)CO_(3)=0.6038` | |
| 102. |
0.16 g a dibasic acid required 25 mL of decinormal NaOH solution for complete neutralisation. The molecular mass of the acid is :A. 32B. 64C. 128D. 256 |
| Answer» Correct Answer - C | |
| 103. |
4 g of a mixture of `Na_(2)SO_(4)` and anhydrous `Na_(2)CO_(3)` were dissolved in pure and volume made up to 250 mL. 20 mL of this solution required 25 mL of `N//5 H_(2)SO_(4)` for complete neutralisation. Calculate the percentage composition of the mixture. |
| Answer» Correct Answer - `Na_(2)CO_(3)=82.8%; Na_(2)SO_(4)=17.2%` | |
| 104. |
10.875 g of a mixture of NaCl and `Na_(2)CO_(3)` was dissolved in water and the volume made up to 250 mL, 20 mL of this solution required 75.5 mL of `(N)/(10) H_(2)SO_(4)`. Find out the percentage composition of the mixture. |
|
Answer» Only `Na_(2)CO_(3)` will react `H_(2)SO_(4)`. Applying `underset((Na_(2)CO_(3)))(N_(1)V_(1)-=underset((H_(2)SO_(4)))(N_(2)V_(2))` `N_(1)xx20=75.5xx(1)/(10)` `N_(1)=(75.5)/(20xx10)=0.3775` `underset(1 " mol.mass")(Na_(2)CO_(3))+underset(2g eq.)(H_(2)SO_(4)) to Na_(2)SO_(4)+H_(2)O+CO_(2)` Eq. mass of `Na_(2)CO_(3)=(106)/(2)=53` Mass of `Na_(2)CO_(3)` present in 250 mL 0.3775 N solution `=(NxxExxV)/(1000)=(0.3775xx53xx250)/(1000)` =5.0018 g Mass of NaCl`=(10.875-5.0018)=5.8732 g` `Na_(2)CO_(3)=(5.0018)/(10.875)xx100=45.99%` `NaCl=(5.8732)/(10.875)xx100=54.0%` |
|
| 105. |
Metallic tin in the presence of HCl is oxidised by `K_(2)Cr_(2)O_(7)` solution to stannic chloride. What volume of decinormal dichromate solution would be reduced by 1 g of Sn ? |
|
Answer» Correct Answer - 337 mL `underset(3xx118.7 g)(3Sn)+underset(2xx294 g)(2K_(2)Cr_(2)O_(7))+28 HCl to 3SnCl_(4)+4KCl+4CrCl_(3)+14H_(2)O` `K_(2)Cr_(2)O_(7)` required for 1 g of Sn`=(2xx294)/(3xx118.7)=1.65 g` |
|
| 106. |
How many grams of `NaHCO_(3)` are required to neutralise 1 mL of 0.0902 N vinegar ?A. `8.4 xx 10^(-3)g`B. `1.5xx10^(-3)g`C. `0.758xx10^(-3)g`D. `1.07xx10^(-3)g` |
|
Answer» Correct Answer - C Number of equivalents of `NaHCO_(3)` =Number of equivalents of acid `=(NV)/(1000)=(0.09202xx1)/(1000)` Mass of `NaHCO_(3)=(0.09202xx1xx84)/(1000)` `=0.758xx10^(-3)g` |
|
| 107. |
`0.7 g " of " Na_(2)CO_(3).xH_(2)O` were dissolved in water and the volume was made to `100 mL, 20 mL` of this solution required `19.8 mL " of " N//10 HCl` for complete neutralization. The value of `x` is:A. 7B. 3C. 2D. 5 |
|
Answer» Correct Answer - C Number of milliequivalents of `Na_(2)CO_(3).xH_(2)O` in 20 mL `=19.8xx(1)/(10)` =1.98 `therefore` Number of milliequivalents in 100 mL `=1.98xx5=9.9` `("Mass")/("Equivalent mass")xx1000=9.9` `(0.7)/(M//2)xx1000=9.9` M=141.40 (Molar mass of `Na_(2)CO_(3). xH_(2)O)` 106+18x=141.40 x=2 |
|
| 108. |
In an ore the only oxidisable material is `Sn^(2+)`. This ore is titrated with a dichromate solution containing `2.5 g K_(2)Cr_(2)O_(7)` in 0.50 litre. A 0.40 g of sample of the ore required `10.0 cm^(3)` of the titrant to reach equivalent point. Calculate the percentage of tin in ore. (K=39.1, Cr=52, Sn=118.7) |
|
Answer» Mol. Mass of `K_(2)Cr_(2)O_(7)=2xx39.1+2xx52+7xx16` =78.2+104.0+112.0 =294.2 Eq. mass of `K_(2)Cr_(2)O_(7)=(294.2)/(6)=49.03` Normality of `K_(2)Cr_(2)O_(7)` solution `=(2.5)/(49.03)xx(1000)/(500)=(5)/(49.03) N` `10 mL (5)/(49.03) N K_(2)Cr_(2)O_(7)-=10 mL (5)/(49.03)N` stannous ion Eq. mass of `Sn^(2+)=(118.7)/(2)=59.35` Amount of Sn in the sample `=(5)/(49.03)xx(59.35)/(1000)xx10` =0.0605 g Percentage of Sn in the ore `=(0.0605)/(0.40)xx100=15` |
|
| 109. |
50 mL sample of ozonised oxygen at NTP was passed through a solution of potassium iodide. The liberated iodine required 15 mL of `0.08 N Na_(2)S_(2)O_(3)` solution for complete titration. Calculate the volume of ozone at NTP in the given sample. |
|
Answer» Reactions involved may be given as : `2KI+H_(2)O+O_(3) to 2KOH+I_(2)+O_(2) uarr` `I_(2)+2Na_(2)S_(2)O_(3) to 2NaI+Na_(2)S_(4)O_(6)` 1 mole `O_(3)=2 mol e Na_(2)S_(2)O_(3)` No. of moles of hypo `=("Mass")/("Molecular mass "(158))` `(ExxNxxV)/(1000xx158)` where, `E_(Na_(2)S_(2)O_(3))=158, N = 0.08, V=15` `therefore` No. of moles of hypo `=(158xx0.08xx15)/(1000xx158)=1.2xx10^(-3)` No. of moles of `O_(3)=(1)/(2)` mole of hypo `=(1)/(2)xx1.2xx10^(-3)` `=6xx10^(-4)` mole Volume of `O_(3)` at NTP =No. of moles`xx 22400` `=6xx10^(-4)xx22400` =13.44 mL at NTP |
|
| 110. |
10 mL of a potassium dichromate solution liberates iodine from potassium iodide solution. When the iodine was titrated with hypo solution `(N//20)`, the titre value was 15 mL. Find the concentration of dichromate solution in g per litre. |
|
Answer» The reaction involved may be given as : `Cr_(2)O_(7)^(2-)+6I^(-)+14H^(+) to 2Cr^(3+)+3I_(2)+7H_(2)O` `3[I_(2)+2Na_(2)S_(2)O_(3)]to 3[2NaI+Na_(2)S_(3)O_(6)]` 1 mole `K_(2)Cr_(2)O_(7)-=6 " moles of" Na_(2)S_(2)O_(3)` No. of moles of hypo `=("Mass")/(" M. w. " (158))=(ExxNxxV)/(1000xx158)` `=7.5xx10^(-4)` mole No. of moles of `K_(2)Cr_(2)O_(7)=(1)/(6)`[No. of moles of `Na_(2)S_(2)O_(3)]` Mass of `K_(2)Cr_(2)O_(7)` in 10 mL solution `=1.25xx10^(-4)xx294` =0.03675 g Weight of `K_(2)Cr_(2)O_(7)` in 10 mL solution `=3.675 g//L` |
|
| 111. |
`0.1 g KIO_(3)` and excess KI when treated with HCl, the iodine is liberated. The liberated iodine required 45 mL solution thiosulphate for titration . The molarity of sodium thisoulphate will be equivalent to :A. 0.252 MB. 0.126 MC. 0.0313 MD. 0.0623 M |
|
Answer» Correct Answer - D The reaction involved are : `IO_(3)^(-)+5I^(-)+6H^(+) to 3I_(2)+H_(2)O` `2Na_(2)S_(2)O_(3)+I_(2) to 2 NaI+Na_(2)S_(4)O_(6)` Number of moles of `I_(2)=3xx " Number of moles of " KIO_(3)` `=(3xx0.1)/(214)` Number of moles of `Na_(2)S_(2)O_(3)=2xx " Number of moles of " I_(2)` `(Mxx45)/(1000)=(2xx3xx0.1)/(241)` M=0.623 |
|
| 112. |
Excess of KI and dil. `H_(2)SO_(4)` were mixed in `50 mL H_(2)O_(2)`. The liberated `I_(2)` required 20 mL of `0.1 N Na_(2)S_(2)O_(3)`. Find out the strength of `H_(2)O_(2)` in g per litre. |
| Answer» Correct Answer - 0.68 g | |
| 113. |
30 mL of `K_(2)Cr_(2)O_(7)` liberated iodine from KI solution when the iodine was titrated with hypo solution `(N//20)`, the titre value was 45 mL. Find the concentration of `K_(2)Cr_(2)O_(7)` in g per litre. |
| Answer» Correct Answer - 3.675 g | |
| 114. |
An excess KI solution is mixed in a solution of `K_(2)Cr_(2)O_(7)` and liberated iodine required 72 mL of 0.05 N `Na_(2)S_(2)O_(3)` for complete reaction. How many grams of `K_(2)Cr_(2)O_(7)` were present in the solution of `K_(2)Cr_(2)O_(7)` ? The reaction occurs as : `Cr_(2)O_(7)^(2-)+6I^(-)+14H^(+) to 2Cr^(3+)+3I_(2)+7H_(2)O` |
|
Answer» The reaction involved may be given as : `Cr_(2)O_(7)^(2-)+6I^(-)+14H^(+) to 2Cr^(3+)+3I_(2)+7H_(2)O` `3[I_(2)+2Na_(2)S_(2)O_(3)]to 3[2NaI+Na_(2)S_(3)O_(6)]` 1 mole `K_(2)Cr_(2)O_(7)-=6 " moles of" Na_(2)S_(2)O_(3)` No. of moles of hypo`=("Mass")/(M.w. (158))=(ExxNxxV)/(1000xx158)` `N_(Na_(2)S_(2)O_(3))=(158xx0.05xx72)/(1000xx158)=3.6xx10^(-3)` No. of moles of `K_(2)Cr_(2)O_(7)=(1)/(6)` [No. of moles of `Na_(2)S_(2)O_(3)`] `=(1)/(6)[3.6xx10^(-3)]=6xx10^(-4)` mole Mass of `K_(2)Cr_(2)O_(7)` in the given solution =No. of moles `xx` Molecular weight `=6xx10^(-4)xx294=0.1764` |
|
| 115. |
`25 mL N K_(2)Cr_(2)O_(7)` acidified solution will liberate .. Iodine from KI solution.A. 0.3175 gB. 3.175 gC. 31.75 gD. 317.5 g |
| Answer» Correct Answer - B | |
| 116. |
To a solution of excess of KI in dilute `H_(2)SO_(4)`. 25 mL of an unknown solution of `KMnO_(4)` were added. The liberated iodine was exactly reduced by 42.5 mL of `N//10 Na_(2)S_(2)O_(3)` solution. Calculate the concentration of `KMnO_(4)` solution. |
| Answer» Correct Answer - `5.372 g//L` | |
| 117. |
0.28 g of a commercial sample of `K_(2)Cr_(2)O_(7)` was dissolved in water. Excess of KI was added to it along with dilute `H_(2)SO_(4)`. Iodine liberated was then titrated against sodium thiosulphate solution containing 24.82 g of `Na_(2)S_(2)O_(3). 5H_(2)O` per litre. The thiosulphate solution required was 50 mL. Find the percentage purity of the sample of `K_(2)Cr_(2)O_(7)`. |
| Answer» Correct Answer - `87.5 %` | |
| 118. |
In an oxidation-reduction, `MnO_(4)^(-)` ion is converted to `Mn^(2+)`, what is the number of equivalents of `KMnO_(4)` (mol. Wt.=158) present in 250 mL of 0.04 M `KMnO_(4)` solution ?A. 0.02B. 0.05C. 0.04D. 0.07 |
|
Answer» Correct Answer - B In redox-reaction : `8H^(+)+MnO_(4)^(-)+5e^(-) to Mn^(2+)+4H_(2)O` Change in oxidation state of `MnO_(4)^(-) =(+7)-(+2)=+5` `therefore N_(KMnO_(4))=M_(KMnO_(4))xx5` `=0.04xx5=0.20` Number of equivalents `=(NV)/(1000)=(0.2xx250)/(1000)=0.05` |
|
| 119. |
A `3.0g` sample containing `Fe_(3)O_(4),Fe_(2)O_(3)` and an inert impure substance is treated with excess of `KI` solution in presence of dilute `H_(2)SO_(4)`. The entire iron is converted to `Fe^(2+)` along with the liberation of iodine. The resulting solution is diluted to `100 mL`. A `20 mL` of dilute solution requires `11.0 mL` of `0.5M Na_(2)S_(2)O_(3)` solution to reduce the iodine present. `A` `50 mL` of the diluted solution, after complete extraction of iodine requires `12.80 mL` of `0.25M KMnO_(4)` solution in dilute `H_(2)SO_(4)` medium for the oxidation of `Fe^(2+)`. Calculate the percentage of `Fe_(2)O_(3)` and `Fe_(3)O_(4)` in the original sample. |
|
Answer» `Fe_(3)O_(4)` is an equimolar mixture of `Fe_(2)O_(3)` and FeO. Thus, the sample contains `Fe_(2)O_(3), FeO` and impurities. The amount of iodine liberated depends on the amount of `Fe_(2)O_(3)` and the entire iron is converted into `Fe^(2+)`. `Fe_(3)O_(4)+2KI+H_(2)SO_(4) to 3FeO+H_(2)O+K_(2)SO_(4)+I_(2)` `Fe_(2)O_(3)+KI+H_(2)SO_(4) to 2FeO+H_(2)O+K_(2)SO_(4)+I_(2)` `5xx11.0 mL " of " 0.5M Na_(2)S_(2)O_(3)-=55.0 mL " of " 0.5N Na_(2)S_(2)O_(3)` soln. `-=55.0 mL " of" 0.5N I_(2)` soln. `-=55.0 mL " of" 0.5N Fe_(2)O_(3)` soln. `=27.5xx10^(-3) " equivalent" Fe_(3)O_(4)` soln. `=13.75xx10^(-3) " moles" Fe_(2)O_(3)` `2xx12.8 mL " of" 0.25 m KMnO_(4)` soln. `-=25.6 mL " of " 1.25 N KMnO_(4)` soln. `-=25.6 mL " of" 1.25N FeO` soln. `=32.0xx10^(-3)` equivalent FeO `=32.0xx10^(-3)` moles FeO Moles of FeO in `Fe_(3)O_(4)=0.032-0.0275=0.0045` Mass of `Fe_(3)O_(4)=0.0045xx232=1.044 g` Moles of `Fe_(2)O_(3)` existing separately =0.01375-0.0045=0.00925 Mass of `Fe_(2)O_(3)=0.00925xx160=1.48 g` `% Fe_(3)O_(4)=(1.044)/(3)xx100=34.8 ` `%Fe_(2)O_(3)=(1.48)/(3)xx100=49.33` |
|
| 120. |
In the reaction, `FeS_(2)+KMnO_(4)+H^(+) to Fe^(3+)+SO_(2)+Mn^(2+)+H_(2)O` the equivalent mass of `FeS_(2)` would be equal to :A. molar massB. `("molar mass")/(10)`C. `("molar mass")/(11)`D. `("molar mass")/(13)` |
|
Answer» Correct Answer - C `Fe^(2+) to Fe^(3+) + e^(-), S_(2)^(2-) to 2S^(4+)+10e^(-)` `therefore FeS_(2) to 2S^(4+)+Fe^(3+)+11e^(-)` Equivalent mass of `FeS_(2) =("Molar mass")/(11)` |
|
| 121. |
A solution of `0.2 g` of a compound containing `Cu^(2+)` and `C_(2)O_(4)^(2-)` ions on titration with `0.02M KMnO_(4)` in presence of `H_(2)SO_(4)` consumes `22.6mL` oxidant. The resulting solution is neutralized by `Na_(2)CO_(3)`, acidified with dilute `CH_(3)COOH` and titrated with excess of `KI`. The liberated `I_(2)` required `11.3 mL "of" 0.05M Na_(2)S_(2)O_(3)` for complete reduction. Find out mole ratio of `Cu^(2+)` and `C_(2)O_(4)^(2+)` in compound. |
|
Answer» 1st case : Only `C_(2)O_(4)^(2-)` ions are oxidised by `KMnO_(4)` solution. Normality of `KMnO_(4)` solution `=0.02xx5=0.1 N` 22.6 mL of 0.1 N `KMnO_(4)=22.6 mL " of" 0.1 N C_(2)O_(4)^(2-)` soln. `underset("in the solution")("Mass of" C_(2)O_(4)^(2-))` ions `=(N xx E xx V)/(1000)=(N xx M xx V)/(1000xx2)` No. of moles of `C_(2)O_(4)^(2-)` ions in the solution`=(N xx M xxV)/(1000xx2xxM)` `=(N xx V)/(2000)` `=(0.1xx22.6)/(2000)` `=11.3xx10^(-4)` 2nd case : Only `Cu^(2+)` ions are reduced by KI and iodine liberated is neutralised by `Na_(2)S_(2)O_(3)` solution. 11.3 mL of `0.05 M Na_(2)S_(2)O_(3)-=11.3 mL of 0.05 N Na_(2)S_(2)O_(3)` `=11.3 mL of 0.05 N I_(2)` `=11.3 mL " of " 0.05 N Cu^(2+)` Mass of `Cu^(2+)` ions in the solution`=(NxxExxV)/(1000)=(NxxMxxV)/(1000)` No. of moles of `Cu^(2+)` ions in the solution `=(N xx M xx V)/(1000xxM)` `=(N xx V)/(1000)` `=(0.05xx11.3)/(1000)` `=5.65xx10^(-4)` Molar ratio of `(Cu^(2+))/(C_(2)O_(4)^(2-))=(5.65xx10^(-4))/(11.3xx10^(-4))=(1)/(2)` |
|
| 122. |
Calculate the equivalent mass of underlines species : 1. `Na_(2)SO_(3)+ul(Na_(2)CrO_(4))to Na_(2)SO_(4)+Cr(OH)_(3)` (ii) `Fe_(3)O_(4)+ul(KMnO_(4)) to Fe_(2)O_(3)+MnO_(2)` (iii) `ul(2Na_(2)S_(2)O_(3))+I_(2)to Na_(2)S_(4)O_(6)+2NaI` (iv) `ul(As_(2)S_(3))+10NO_(3)^(-)+4H^(+) to 10NO_(2)+2AsO_(4)^(3-)+3S+2H_(2)O` (v) `ul(H_(3)PO_(3)) to H_(3)PO_(4)+PH_(3)` (vi) `ul(5SO_(2))+2KMnO_(4)+2H_(2)O to K_(2)SO_(4)+2MnSO_(4)+2H_(2)_(2)SO_(4)` |
| Answer» Correct Answer - (i) 54; (ii) 52.66; (iii) 158; (iv) `M.w.//10;` (v) 41; (vi) 32 | |
| 123. |
A sample of coconut oil weighing 1.5763 g is mixed with 25 mL of 0.4210 M KOH. Some KOH is used in saponification of coconut oil. After the saponification is complete, 8.46 mL of `0.2732 M H_(2)SO_(4)` is required to neutralize excess KOH. The saponification number of peanut oil is :A. 209.6B. 98.9C. 108.9D. 218.9 |
|
Answer» Correct Answer - A Number of milliequivalent of KOH added`=25xx0.421=10.525` Number of milliequivalents left unreacted =Number of milliequivalents of `H_(2)SO_(4)` used `= 8.46xx0.2732xx24.623 " " ("Here, basicity of " H_(2)SO_(4)=2)` Number of milliequivalents of KOH used by oil =10.525-4.623=5.902 Mass of KOH used `=(5.902xx56)/(1000)=0.3305 g =330.5 mg` Saponification number = Mass of KOH in mg used by 1 g oil or fat `=(0.3305xx1000)/(1.5763)=209.6` |
|
| 124. |
In the reaction, `I_(2)+2S_(2)O_(3)^(2-) to 2I^(-)+S_(4)O_(6)^(2-)` equivalent mass of iodine is :A. equal to its molecular massB. `1//2` the molecular massC. `1//4` the molecular massD. twice the molecular mass |
| Answer» Correct Answer - B | |
| 125. |
In the reaction, `I_(2)+2S_(2)O_(3)^(2-) to 2I^(-)+S_(4)O_(6)^(2-)` equivalent weight will be equal to :A. `4//6` of molecular weightB. molecular weightC. `2//9` of molecular weightD. twice the molecular weight |
| Answer» Correct Answer - B | |
| 126. |
Assertion : In the reactin, `2S_(2)O_(3)^(2-)+I_(2) to S_(4)O_(6)^(2-) + 2I^(-): I_(2)` is oxidised. Reason : During oxidation, loss of electron takes place.A. If both Assertion and Reason are correct, and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.C. If Assertion is correct, but Reason is incorrect.D. If Assertion is incorrect, but Reason is correct. |
| Answer» Correct Answer - D | |
| 127. |
Chemists work with standardised solution, a solution whose concentration is known. The requirements to standardise the solution are : 1. the volume of the solution. 2. the number of moles of solute in that volume. A primary standard solution is used in determining the molarity of a solution. To find the molarity of HCl, 0.317 g of `Na_(2)CO_(3)`, the primary standard dissolved in water, is used in titrating the solution of HCl. 22.9 mL of acid are required to neutralise the sodium carbonate. This is the needed volume (first requirement). The stoichiometric equation is used to know the second requirement. `2HCl(aq.)+Na_(2)CO_(3)(aq.) to 2NaCl+H_(2)O+CO_(2) uarr` What is the molarity of HCl in the above case ?A. 0.261 MB. 0.522 MC. 0.1 MD. 1 M |
| Answer» Correct Answer - A | |
| 128. |
Oxalic acid dihydrate, `H_(2)C_(2)O_(4). 2H_(2)O(s)` is often used as a primary reagent to standardise sodium hydroxide solution. Which of these facts are reasons to choose this substance as a primary standard ? I. It is diprotic. II. It is a stable compound that can be weighed directly in air. III. It is available in pure form.A. III onlyB. I and II onlyC. II and III onlyD. I, II and III |
| Answer» Correct Answer - C | |
| 129. |
Chemists work with standardised solution, a solution whose concentration is known. The requirements to standardise the solution are : 1. the volume of the solution. 2. the number of moles of solute in that volume. A primary standard solution is used in determining the molarity of a solution. To find the molarity of HCl, 0.317 g of `Na_(2)CO_(3)`, the primary standard dissolved in water, is used in titrating the solution of HCl. 22.9 mL of acid are required to neutralise the sodium carbonate. This is the needed volume (first requirement). The stoichiometric equation is used to know the second requirement. `2HCl(aq.)+Na_(2)CO_(3)(aq.) to 2NaCl+H_(2)O+CO_(2) uarr` Equivalent mass of `Na_(2)CO_(3)` in the above equation will be :A. 106B. 53C. 26.5D. 13.25 |
| Answer» Correct Answer - B | |
| 130. |
Chemists work with standardised solution, a solution whose concentration is known. The requirements to standardise the solution are : 1. the volume of the solution. 2. the number of moles of solute in that volume. A primary standard solution is used in determining the molarity of a solution. To find the molarity of HCl, 0.317 g of `Na_(2)CO_(3)`, the primary standard dissolved in water, is used in titrating the solution of HCl. 22.9 mL of acid are required to neutralise the sodium carbonate. This is the needed volume (first requirement). The stoichiometric equation is used to know the second requirement. `2HCl(aq.)+Na_(2)CO_(3)(aq.) to 2NaCl+H_(2)O+CO_(2) uarr` The suitable indicator in the above titration will be :A. phenolphthaleinB. methyl orangeC. litmusD. bromothymol blue |
| Answer» Correct Answer - B | |
| 131. |
The value of n in the equation, `Cr_(2)O_(7)^(2-) +14 H^(+)+ne^(-) to 2Cr^(3+) + 7H_(2)O` is :A. 2B. 3C. 4D. 6 |
| Answer» Correct Answer - D | |
| 132. |
The equivalent weight of `KIO_(3)` in the reaction, `2Cr(OH)_(3)+OH^(-)+KIO_(3) to 2CrO_(4)^(2-) + 5H_(2)O + KI` is :A. molecular weightB. `("molecular weight ")/(3)`C. `("molecular weight ")/(6)`D. `("molecular weight")/(2)` |
| Answer» Correct Answer - C | |
| 133. |
In 20 mL of a solution of HCl, 3 g of `CaCO_(3)` were dissolved, 0.5 g of `CaCO_(3)` being left undissolved. Find out the strength of this solution in terms of (i) normality and (ii) `g//L`. Find the volume of this acid which would be required to make 1 litre of normal solution of this acid. |
| Answer» Correct Answer - `91.25 g//L, 2.52 N, 400 mL` | |
| 134. |
Acidified `KMnO_(4)` oxidizes acid to `CO_(2)`. What is the volume ( in litre) of `10^(-4) M KMnO_(4)` required to completely oxidize 0.5 litre of `10^(-2)` M oxalic acid in acid medium ?A. 125B. 1250C. 200D. 20 |
|
Answer» Correct Answer - D `2KMnO_(4)+3H_(2)SO_(4)+5{:(COOH,),(|" ",),(COOH,):}toK_(2)SO_(4)+2MnSO_(4)+10CO_(2)+8H_(2)O` `(M_(1)V_(1))/(n_(1))(KMnO_(4))=(M_(2)V_(2))/(n_(2)){:((COOH,),(|" ",),(COOH,)):}` `(10^(-4)xxV_(1))/(2)=(10^(-2)xx0.5)/(5)` `V_(1)=20 L` |
|
| 135. |
One kilogram of sea water contains 6 mg of dissolved `O_(2)`. The concentration of `O_(2)` in the sample in ppm is :A. 0.6B. `6.0`C. `60.0`D. `2.0` |
|
Answer» Correct Answer - B 1 kg water `=10^(6)` mg: `10^(6)` mg water contains 6 mg `O_(2)` `therefore` Concentration of `O_(2)` is 6 ppm. |
|