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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.
| 151. |
When burnt in air, a 12.0 g mixture of carbon and sulphur yields a mixture of `CO_2` and `SO_2` , in which the number of moles of `SO_2` is half that of `CO_2`.The mass of the carbon the mixture contains is : (At . Wt. S=32)A. 0.75B. 0.5C. 0.4D. 0.25 |
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Answer» Correct Answer - B Let weight of C be xg, then S will be `(14 -x)g` `(x//12)/((14-x)//32) = (2)/(1)` `:. X = 6g`, moles of `C = (6)/(12) = 0.5` |
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| 152. |
If 5ml of methane is completely burnt the volume of oxygen required and the volume of `CO_(2)` formed under the same conditions areA. 5ml, 10mlB. 10 ml, 5mlC. 5ml, 15mlD. 10 ml, 10ml |
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Answer» Correct Answer - B `CH_(4) +2O_(2) rarr CO_(2) +2H_(2)O` |
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| 153. |
A compound contains `92.3%` of carbon and `7.7%` of hydrogen. The moleculer of the compound is 39 times heavier than hydrogen molecule. The molecular formula of the compound isA. `C_(3)H_(3)`B. `C_(2)H_(2)`C. `C_(2)H_(4)`D. `C_(6)H_(6)` |
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Answer» Correct Answer - D `GMW = 2 xx 39 = 78 g(GMW_(H_(2))=29)` `M.F = E.Fxx (GMW)/(EFW)` |
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| 154. |
In a quality control analysis for sulphur impurity `0.56g` steel sample was burnt in a stream of oxygen and sulpphur was converted into `SO_(2)` gas. The `SO_(2)` was then oxidized to sulphate by using `H_(2)O_(2)` solution to which had been added `30mL` of `0.04M NaOH`. the equation for reaction is: `SO_(2(g)) +H_(2)O_(2(aq)) +2OH_((aq))^(-) rarr SO_(4(aq))^(-2) +2H_(2)O_((1))` `22.48 mL` of `0.024M HCI` was required to neutralize the base remaining after oxidation reaction. Calculate % of sulphur in given sample.A. `3.6%`B. `1.875%`C. `9%`D. `4.5%` |
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Answer» Correct Answer - B Meq of alkali added `=30 xx 0.04 = 1.2` Meq. Of alkali left `=22.48 xx 0.024 = 0.54` `:.` Meq. of alkali for `SO_(2)` and `H_(2)O_(2) = 1.2 - 0.54 = 0.66 =` M.eqts of alkali 1 moles of alkai `-=1` moles of `SO_(2)` `:.` Weight of `S = (0.66)/(2) xx (32)/(1000) = 0.0105g` `:. %` of `S = (0.0105)/(0.56) xx 100 = 1.875%` |
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| 155. |
20 ml of the solution containg `Na_(2)CO_(3)` and `NaHCO_(3)` is titrated with `0.1MHCI` using Phenolphthalein indicator the end point was `10ml. 20ml` of the same solution is titrated with `0.1M HCI`, the end point was 25ml with Methylorange indicaor from the begining. `NaHCO_(3)+HCl rarr NaCl +H_(2)O +CO_(2)` `Na_(2)CO_(3)+2HCl rarr 2NaCl +CO_(2) +H_(2)O` What amount of `NaOH` is required to convert `NaHCO_(3)` to `Na_(2)CO_(3)` in 1 litre of solutionA. `2g`B. `20g`C. `1g`D. `0.5g` |
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Answer» Correct Answer - C `NaHCO_(3)+NaOH rarr Na_(2)CO_(3) +H_(2)O` 84g of `NaHCO_(3)` requires 40g of `NaOH` `2.1g` of `NaHCO_(3)` requires? `= (40)/(84) xx 2.1 = 1g` |
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| 156. |
20 ml of the solution containg `Na_(2)CO_(3)` and `NaHCO_(3)` is titrated with `0.1MHCI` using Phenolphthalein indicator the end point was `10ml. 20ml` of the same solution is titrated with `0.1M HCI`, the end point was 25ml with Methylorange indicaor from the begining. `NaHCO_(3)+HCl rarr NaCl +H_(2)O +CO_(2)` `Na_(2)CO_(3)+2HCl rarr 2NaCl +CO_(2) +H_(2)O` What is amount of `NaHCO_(3)` present in 1 litre of solutionA. `2.1g`B. `1.05g`C. `8.1g`D. `0.844g` |
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Answer» Correct Answer - A Using methyl orange, `x +y = 25 xx 0.1 = 2.5, y = 2.5 - 2 = 0.5` Wt of `NaHCO_(3)` in 20ml `=(0.5)/(1000) xx 84 = 0.042g` wt of `NaHCO_(3)` in 1 lit `=(0.042)/(20) xx 1000 = 2.1g` |
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| 157. |
A `0.518 g` sample of limestone is dissolved in `HCl` and then the calcium is precipitated as `CaC_(2)O_(4)`. After filtering and washing the precipitate, it requires `40.0` filtering and washing the precipitate, it requires `40.0 mL` of `0.250 N KMnO_(4)`, solution acidified with `H_(2)SO_(4)` to titrate it as. The percentage fo `CaO` in the sample is: `MnO_(4)^(-)+H^(+)+C_(2)O_(4)^(2-)rarrMn^(2+)+CO_(2)+2H_(2)O`A. `54.0%`B. `27.1%`C. `42%`D. `84%` |
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Answer» Correct Answer - A No of milli equivalents of lime stone = no of milli equivalents of `CaC_(2)O_(4) =` no of milli equivalents of `KMnO_(4) = 40 xx 0.25 = 10` No of milli moles of lime stone = No of milli moles of lime =5 `:.` weight of `CaO = 5 xx 56 xx 10^(-3) = 0.28g` percentage of `CaO = (0.28)/(0.518) xx 100 = 54.05%` |
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| 158. |
21.6 g of silver coin is dissolved in `HNO_(3)`. When NaCl is added to this solution, all silver is precipitated as AgCl. The weight of AgCl is found to be 14.35g then % silver in coin is: `Ag + HNO_(3) overset(NaCl)to AgCl`A. `50%`B. `75%`C. `100%`D. `15%` |
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Answer» Correct Answer - A `NaCl +AgNO_(3) rarr AgCl +naNO_(3)` |
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| 159. |
Quantity of 1.5g brass containing Cu,Zn reacts with `3M,HNO_(3)` solution, the following reactions takes place `Cu +HNO_(3) rarr Cu^(2+) +NO_(2) +H_(2)O` `Zn +H^(+) +NO_(3)^(-) rarr NH_(4)^(+) +ZN^(2+) +H_(2)O` (un balanced) The liberated `NO_(2)(g)` is found to occupy 1.04 litre at `25^(@)C` and 1 atm pressure. Now answer the following question % of copper in the alloy isA. `9.56`B. `14.34`C. `6.37`D. `19.12` |
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Answer» Correct Answer - B `Cu + 4HNO_(3) rarr Cu (NO_(3))_(2) +2NO_(2) +2H_(2)O` `4Zn + 10 HNO_(3) rarr NH_(4)NO_(3) +4Zn (NO_(3))_(2) +3H_(2)O` |
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| 160. |
`50 gr` of sample of sodium hydroxide required for complete neutralisation, 1 litre `1 NHCl`. What is the percentage purity of `NaOH` isA. 50B. 60C. 70D. 80 |
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Answer» Correct Answer - D `w = N xx GEW xx V` (in lit) % of purity of `NaOH = (40)/(50) xx 100 = 80%` |
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| 161. |
150 ml `(M)/(10) Ba(MnO_(4))_(2)` in acidic medium can oxidise completelyA. `150mL 1MFe^(+2)`B. `50mL 1M FeC_(2)O_(4)`C. `75mL 1M C_(2)O_(4)^(-2)`D. `25mL 1M K_(2)Cr_(2)O_(7)` solution |
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Answer» Correct Answer - A::B::C::D No. of m.eqts of `Ba(MnO_(4))_(2)` reacted `=150 xx (1)/(10) xx 10 = 150` |
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| 162. |
The volume occupied by 35.5g of `CI_(2)` at S.T.P. |
| Answer» `=(35.5)/(71) xx 22.4 = 11.2` lit. | |
| 163. |
`V_(1)` ml of NaOH of molarity X and `V_(2)` ml of `Ba(OH)_(2)` of molarity `(y)/(2)` are mixed together. Mixture is completely neutralized by 100 ml `(0.1)/(2)"M H"_(2)SO_(4)`/. If `(V_(1))/(V_(2))=(1)/(4)` and `(x)/(y)=4`, what fraction of acid is neutralized by `Ba(OH)_(2)`?A. `0.5`B. `0.25`C. `0.33`D. `0.67` |
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Answer» Correct Answer - D `NaOH = XV_(1)` total no. of milli moles of acid used `=100 xx 0.1 = 10` Let K milli moles of acid used for neutralisation `Ba(OH)_(2) :. 2YV_(2) = K` milli moles of acid used for neutralising `NaOH :. x V_(1) = (10-K)` Fraction of acid used for neutralising `Ba(OH)_(2)` `= (K)/(K+(10-K)) = =(2YV_(2))/(XV_(1)+2YV_(2))` given that `(X)/(y) = 4, (V_(1))/(V_(2)) = (1)/(4)` `rArr X = 4Y, V_(1) = (V_(2))/(4)` Fraction of acid used for neutralising `Ba(OH)_(2)` `=(2YV_(2))/(4Yxx (V_(2))/(4)+2YV_(2)) = (2)/(3) = 0.67` |
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| 164. |
A solution containing `0.05` mole of `SO_(2)CI_(2)` in water will be completely neutralized byA. 50ml of `4.0M NaOH`B. 25ml of `0.2M KMnO_(4)`C. 50ml of `0.5M KOH`D. 50ml of `1M KOH` |
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Answer» Correct Answer - A `SO_(2)Cl_(2) +2H_(2)O rarr H_(2)SO_(4)+2HCl` |
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| 165. |
`12.5 mL` of a solution containing `6.0 g` of a dibasic acid in `1 L` was found to be neutralized by `10 mL` a decinormal solution of `NaOH`. The molecular weight of the acid isA. 75B. 110C. 120D. 150 |
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Answer» Correct Answer - D `N = (w)/(GEW) xx (1000)/(V("in ml"))` |
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| 166. |
0.16 g of dibasic acid required 25 ml of decinormal NaOH solution for complete neutralisation. The molecular weight of the acid will beA. 32B. 64C. 128D. 256 |
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Answer» Correct Answer - C Ratio of molecules is equal to inverse ratio of their mol. Wts. If the wts are equal |
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| 167. |
Three different solution of oxidising agents. `K_(2)Cr_(2)O_(7),I_(2), and KMnO_(4)` is titrated separately with `0.19g of K_(2)S_(2)O_(3)`. The molarity of each oxidising agent is 0.1 M and the reaction are: (i). `Cr_(2)O_(7)^(2-)+S_(2)O_(3)^(2-)toCr^(3+)+SO_(4)^(2-)` (ii). `I_(2)+S_(2)O_(3)^(2-)toI^(ɵ)+S_(4)O_(6)^(2-)` (iii). `MnO_(4)^(ɵ)+S_(2)O_(3)^(2-)toMnO_(2)+SO_(4)^(2-)` (Molecular weight of `K_(2)S_(2)O_(3)=190,K_(2)Cr_(2)O_(7)=294,KMnO_(4)=158, and I_(2)=254 mol^(-1))` Which of the following statements is/are correct?A. All three oxidants can act as self-indicatorsB. Volume of `I_(2)` used is miniumC. Volume of `K_(2)Cr_(2)O_(7)` used is maximumD. Wt. of `KMnO_(4)` used in the titration is max |
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Answer» Correct Answer - A::B::D All three are self indicators i.e, they do not need any indicator for titration. `K_(2)S_(2)O_(3) = (0.19)/(190) = 10^(-3)mol` = 1 mmol (i) mEq of `S_(2)O_(3)^(2-) (n = 8) = ` mEq of `Cr_(2)O_(7)^(2-) (n = 6), 1 xx 8 = 0.1 xx 6 xx V` `:. V_(Cr_(2)O_(7)^(2-)) = (80)/(6)mL` (ii) mEq of `S_(2)O_(3)^(2-) (n = 1)=` mEq of `I_(2) (n = 2)` `1 xx 1 = 0.1 xx 2 xx V :. V_(I_(2)) = 5 mL` (iii) mEq of `S_(2)O_(3)^(2-) (n = 8)` = mEq of `MnO_(4)^(-) (n = 3) , 1xx 8 = 0.1 xx 3 xx V` `:. V_(MnO_(4)^(-)) = (80)/(3)mL` Also, Weight of `K_(2)Cr_(2)O_(7) = 8 xx (294)/(6) xx 10^(-3) = 0.392g` Weight of `I_(2) = 1 xx (254)/(2) xx 10^(-3) = 0.127g` Weight of `KMnO_(4) = 8 xx (158)/(3) xx 10^(-3) = 1.264g` |
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| 168. |
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. Statement -1 is True, Statement -2 is True, Statement -2 is a correctB. Statement -1 is True, Statement -2 is True, statement -2 is NOT a correct explanation for Statement-8C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
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Answer» Correct Answer - A `overset(2+)(Fe)overset(1-)(S_(2)) overset(O_(2))rarr overset(3+)(Fe_(2))O_(3) +overset(4+)(SO_(2))` |
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| 169. |
`x g` of `H_2O_2` requires `100mL` of `M//5 KMnO_4` in a titration in a solution having `pOH = 1.0` Which of the following is`//`are correct?A. The value of x is `1.7g`B. The value of x is `0.34g`C. `MnO_(4)^(-)` changes to `MnO_(4)^(2-)`D. `H_(2)O_(2)` changes to `O_(2)` |
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Answer» Correct Answer - B::C::D `pOH =1`, strong basic medium `rArr MnO_(4)^(-) rarr MnO_(4)^(2-)` (n-factor =1) `2bar(O)H +H_(2)O_(2) rarr O_(2)+2H_(2)O +2e^(-)` meq of `H_(2)O_(2) =` meq of `MnO_(4)^(-)` `(x)/(34//2) xx 1000 = 100 xx ((1)/(5)xx1), x = 0.34g` |
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| 170. |
Atomicity of oleum `(H_(2)S_(2)O_(7))` isA. 11B. 8C. 7D. 18 |
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Answer» Correct Answer - A `H_(2)S_(2)O_(7)` contains 11 atom. |
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| 171. |
In this reaction: `S_(2)O_(8)^(2-)+2I^(-) to 2SO_(4)^(2-)+I_(2)`A. Oxidation of iodide into iodine takes placeB. Reduction of iodine into iodide takes placeC. Both oxidation ane reduction of iodine takes placeD. None of the above |
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Answer» Correct Answer - A `2I^(-) rarr overset(0)I_(2)` |
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| 172. |
Which of the following is/are correct about the redox reation? `MnO_(4)^(-) +S_(2)O_(3)^(2-) +H^(o+) rarr Mn^(2+) +S_(4)O_(6)^(2-)`A. 1 mol of `S_(2)O_(3)^(2-)` is oxidised by 8 mol of `MnO_(4)^(-)`B. The above redox reaction with the change of pH from 4 to 10 will have and effect on the stiochiometry of the reactionC. Change of pH from 4 to 7 will change the nature of the productD. At `pH = 7, S_(20O_(3)^(2-)` ions are oxidised to `HSO_(4)^(-)` |
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Answer» Correct Answer - B::C::D (a) `5e^(-) + MnO_(4)^(-) rarr Mn^(2+) (n= 5)` `2S_(2)O_(3)^(-) rarr S_(4)O_(6)^(2-) +2e^(-) (n = (2)/(2) = 1)` Eq of `MnO_(4)^(-) -=` Eq of `S_(2)O_(3)^(-)` `5xx `moles of `MnO_(4)^(-) -= 1 xx` moles of `S_(2)O_(3)^(2-)` `:. 1` mol of `S_(2)O_(3)^(2-) = 5` mol of `MnO_(4)^(-)` (b) pH changes from 4 to 10 (acidic to strongly basix) `e^(-) + MnO_(4)^(-) rarr MnO_(4)^(2-) (n = 1)` `S_(2)O_(3)^(2-) rarr 2SO_(4)^(2-) +8e^(-) (n = 8)` Eq of `MnO_(4)^(-) =` Eq of `S_(2)O_(3)^(-)` `:. 1` mol of `S_(2)O_(3)^(2-) = (1)/(8)` mol of `MnO_(4)^(-)` Hence with change of pH from 4 to 10, will change the stiochiometry of reaction and alo changes the product. (c) pH changes from 4 to 7 (acidic to neutral medium) `3e^(+) +MnO_(4)^(-) rarr MnO_(2) (n = 3)` `S_(2)O_(3)^(2-) rarr 2HSO_(4)^(-) +8e^(-) (n = 8)` Hence it will also effect the stoichiometry of reaction and nature of product. (d) At `pH = 7, S_(2)O_(3)^(2-)` is oxidised to `HSO_(4)^(-)` ion |
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| 173. |
Statement-1 : The percentage of nitrogen in Urea is approximately `46.6%`. Statement-2 : Urea is an ionic compound.A. Statement -1 is True, Statement -2 is True, Statement -2 is a correctB. Statement -1 is True, Statement -2 is True, statement -2 is NOT a correct explanation for Statement-2C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
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Answer» Correct Answer - B Urea is `H_(2)NCONH_(2)%` of `N =(28)/(60) 100 = 46.6%` Urea is a colvent compound |
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| 174. |
`A` spherical ball of radius `7cm` contains `56%` iron. If density is `1.4 g//cm^(3)`, number of mol of `Fe` present approximately is :A. 10B. 15C. 20D. 25 |
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Answer» Correct Answer - C Mass of spherical ball `= (4)/(3) xx (22)/(7) xx 7^(3) xx 1.4` `= 2012.26g` no. of moles of `Fe = (56)/(100) xx (2012)/(56) = 20.12` |
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| 175. |
The number of atoms in 4.25 g of `NH_(3)` is approximatelyA. `1 xx 10^(23)`B. `1.5 xx 10^(23)`C. `2 xx 10^(23)`D. `6 xx 10^(23)` |
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Answer» Correct Answer - D `(4.25)/(17) xx 4 = 1` |
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| 176. |
The equilibrium constant of the following redox rection at 298 K is `1 xx 10^(8)` `2Fe^(3+) (aq.) +2I^(-) (aq.) hArr 2Fe^(2+)(aq.) +I_(2) (s)` If the standard reducing potential of iodine becoming iodide is +0.54 V. what is the standard reduction potential of `Fe^(3+)//Fe^(2+)` ?A. `64.2`B. `51.0`C. `48.4`D. `25.5` |
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Answer» Correct Answer - B No. of equivalent of copper (II) salt = No. of equivalent of Iodine =No. of equivalents of hypo `rArr` No. of equivalent of copper (II) salt `= 24.5 xx 0.1 = 2.45` No. of milli moles of copper (II) salt `=2.45` Weight of copper in copper (II) salt `=2.45 xx 63.5 xx 10^(-3) = 0.1556g` `0.305g` of copper (II) salt contain `=0.1556g` of Cu `:. %` copper `=(0.1556)/(0.305) xx 100 = 51%` |
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| 177. |
`H_(2)O_(2)` solution used for hair bleaching is sold as a solution of approximately 5.0 g `H_(2)O_(2)` per 100 mL of the solution. The molecular mass of `H_(2)O_(2)` is 34. The molarity of this solution is approximatelyA. 0.15MB. 1.5MC. 3.0MD. 3.4M |
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Answer» Correct Answer - B `M = (w)/(GMW) xx (1000)/(V "in ml")` |
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| 178. |
Which one of the following is an example of disporoportionation reaction ?A. `3Cl_(2) (g) +6OH^(-) (aq) rarr ClO_(3)^(-) (aq) +5CI^(-) (aq) +3H_(2)O(1)`B. `Ag^(2+) (aq) +Ag (s) rarr 2Ag^(-4) (aq)`C. `Zn(s) +CuSO_(4) (aq) rarr Cu(s) +ZNSO_(4)(aq)`D. `2KClO_(3)(s) rarr 2KCl (s)+ 3O_(2)(g)` |
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Answer» Correct Answer - A `3overset(0)(Cl_(2)) (g)+6OH^(-)(aq)` `rarr ClO_(3)^(-) (aq) +5CI^(-)(aq) +3H_(2)O(l)` |
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| 179. |
The percentage of hydrogen in water and hydrogen peroxide is 11 .1 and 5.9 respectively . These figures illustrateA. Constant ProportionsB. Conservation of massC. Multiple ProportionD. Law of Gaseous volume |
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Answer» Correct Answer - C Law of multiple proportions |
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| 180. |
Statement-1 : Molarity of a solution and molality of a solution both change with density. Statement-2 : Density of the solution changes when percentage by mass of solution changes.A. Statement -1 is True, Statement -2 is True, Statement -2 is a correctB. Statement -1 is True, Statement -2 is True, statement -2 is NOT a correct explanation for Statement-3C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
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Answer» Correct Answer - A `M = (10Dx)/(M_(s))` where `D =` density of solution `x = %` by mass, `M_(s) =` molar mass of solute: |
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| 181. |
Law of multiple proportion is illustrated by:A. Sodium chloride and sodium bromideB. Ordinary water and heavy waterC. Caustic soda and caustic potashD. Sulphur dioxide and sulphur trioxide |
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Answer» Correct Answer - D `SO_(2)` & `SO_(3)` illustrate law of multiple proportions |
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| 182. |
The law of multiple proportions was observed for the pairA. `CO_(2),NO_(2)`B. `N_(2)O,NO_(2)`C. `H_(2)O,H_(2)S`D. `H_(2)S,SO_(2)` |
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Answer» Correct Answer - B `N_(2)O` and `NO_(2)` follows law of multiple proportion |
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| 183. |
The composition of compound A is `40%` X and `60Y`. The composition of compound B is `25%` X and `75%Y`. According to the law of multiple Proportions the ratio of the weight of element Y in compounds A and B is:A. `1:2`B. `2:1`C. `2:3`D. `3:4` |
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Answer» Correct Answer - A `{:(,X,Y),(Ararr,1,(60)/(40)),(Brarr,1,(75)/(25)):}` |
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| 184. |
The density of `3M` solution of `NaCl` is `1.25 g mL^(-1)`. The molality of the solution isA. `2.79 m`B. `1.79 m`C. `3.5m`D. `5.58m` |
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Answer» Correct Answer - A `m = (M xx 1000)/((1000 xx d)-(M xx G.MW))` `N_(1)V_(1) = N_(2)V_(2)` strength `=N xx EW` `MW = EW xx` Basicity `m = (M xx 1000)/(1000 xx d - M xx GMW)` |
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| 185. |
If dry air is considered to have only nitrogen and oxygen and is found to have average molecular mass 28.84. what is the composition of dry Air ? |
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Answer» `28.84 = (28 xx x + 32(100 -x))/(100)` `x = 0.79` i.e `N_(2) = 79%, O_(2) = 21%` |
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| 186. |
The molarity of a solution obtained by mixing 750 mL of 0.5 M HCl with 250 mL of 2 M HCl will beA. `1.00M`B. `1.75M`C. `0.975M`D. `0.875M` |
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Answer» Correct Answer - D `M = (M_(1)V_(1) +M_(2)V_(2))/(V)` |
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| 187. |
Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution.A. 0.2B. 0.4C. 0.1D. 0.25 |
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Answer» Correct Answer - B `M = (w)/(GMW) xx (1000)/(V "in ml")` |
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| 188. |
The density of 3 molal solution of NaOH is 1.110g `mL^(-1)`. Calculate the molarity of the solution.A. 2MB. 3.5MC. 1.5MD. 3M |
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Answer» Correct Answer - D 3 molal solution of `NaOH` means that 3 mole of `NaOH` are dissolved in 1000g of solvent. `:.` Mass of solution = Mass of solvent `+` Mass of solute `=1000g + (3 xx 40g) = 1120g` Volume of solution `=(1120)/(1.110) mL = 1009.00mL` (Since density of solution `= 1.110 gmL^(-1))` Since 1009 mL solution contains 3 moles of `NaOH` `:.` Molarity `=("Number of moles of solute")/("Volume of solution in litre")` `=(3 mol)/(1009.00) xx 1000 = 2.97 M ~~ 3M` |
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| 189. |
40 mg of pure sodium hydroxide is dissolved in 10 L of distilled water. The pH of the solution isA. 5 gmB. 2gC. 12.5gD. 4 gm |
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Answer» Correct Answer - C `w = (M xx GMW xx V("in ml"))/(1000)` |
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| 190. |
100 ml of ethylalcohol is made upon a litre with distilled water. If the density of `C_(2)H_(5)OH` is `0.46gr//ml`. Then its molarity isA. `0.55 m`B. `111m`C. `2.22m`D. `3.33m` |
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Answer» Correct Answer - B `m = (w)/(GMW) xx (1000)/("Wt. of solvent"(g))` |
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| 191. |
Volume of a gas at STP is `1.12xx10^(-7)` c c. Calculate the number of molecules in itA. `6.02 xx 10^(23)`B. `3.01 xx 10^(12)`C. `6.02 xx 10^(12)`D. `3.01 xx 10^(23)` |
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Answer» Correct Answer - B `22400` cc of a gas at `STP = 6.02 xx 10^(23)` molecules `1.2 xx 10^(-7)c c` of a gas at STP = ? `= (1.12 xx 10^(-7) xx 6.02 xx 10^(23))/(22400) = 3.01 xx 10^(12)` molecules |
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| 192. |
Calculate the number of molecules in 1120 ml of hydrogen at STP? |
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Answer» number of molecules in 22400 ml of hydrogen at STP `= 6.023 xx 10^(23)` Number of molecules in 1120 ml of hydrogen `= (1120 xx 6.023 xx 10^(23))/(22400) = 0.3011 xx 10^(23)` |
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| 193. |
`10cm^(3)` of 0.1N monobasic acid requires `15cm^(3)` of sodium hydroxide solution whose normality isA. `0.066N`B. `0.66N`C. `1.5N`D. `0.15N` |
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Answer» Correct Answer - A `N_(1)V_(1) = N_(2)V_(2)` |
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| 194. |
Statement Addition of water to a solution containing solute and solvent changes its normality or molarity only. Explanation The milli-equivalent and milli-moles of solutes are not changed on dilution.A. Statement -1 is True, Statement -2 is True, Statement -2 is a correctB. Statement -1 is True, Statement -2 is True, statement -2 is NOT a correct explanation for Statement-5C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
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Answer» Correct Answer - A meq `=N xx V` (ml) `= (wt)/(Eq.wt) xx 100`. No doubt N decrease with dilution but V increases and thus me. Remain constant |
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| 195. |
The number of molecules present in one milli litre of a gas at STP is known asA. Avogadro numberB. Boltzman numberC. Loschmidt numberD. Methane |
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Answer» Correct Answer - C Loschmidt number is no of molecules in one mL of a gas under STP. |
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| 196. |
The reaction `H_(3)PO_(4)+Ca(OH)_(2)rarrCa(HPO_(4))_(2)+2H_(2)O` Which statements `(s)` is (are)true?A. the equivalent weight of `H_(3)PO_(4)` is 49B. the resulting solution is neutralized by 1 mole of `KOH`C. 1 mole of `H_(3)PO_(4)` is completely neutralized by `1.5` mole of `Ca(OH)_(2)`D. 1 mole of `H_(3)PO_(4)` can be neutralised completely by 1 mole of `Ca(OH)_(2)` |
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Answer» Correct Answer - A::B::C `H_(3)PO_(4) + Ca(OH)_(2) rarr CaHPO_(4)+2H_(2)O` n-factor for `H_(3)PO_(4) =2` (since `2H^(+)` ions are replaced) eq. wt `=(M)/(2) = (98)/(2) = 49` resulting solution of `CaHPO_(4)` have only are replaceable `H^(+)` so nf =1 hence no of eq = 1 for more so can be neutralized by 1 mole of KOH eq of `CaHPO_(4) =` eq. of KOH For complete neutralization-no. of eq. of `H_(3)PO_(4) =` no. of eq. of `Ca(OH)_(2)` `1 xx 3 = 1.5 xx 2` can be neutralized. |
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| 197. |
Which of the following has the highest normality ? ( consider each of the acid is 100% ionised )A. 8 gr KOH per `100ml`B. `0.5 M H_(2)SO_(4)`C. 6gr of `NaOH` per 100 mlD. `1N H_(3)PO_(4)` |
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Answer» Correct Answer - C `N = (w)/(GEW) xx (1000)/(V("in ml"))` `N = Mxx` Basicity of an acid |
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| 198. |
Two students performed the same experiment separately and each one of them recovered two readings of mass which are given below. Correct reading of mass is 3.0 g. On the basis of given data, mark the correct optioin out of the following statements. A. Results of both the students are neither accurate nor preciseB. Results of student A are both precise and accurateC. Results of student B are neither precise nor accurateD. Results of student B are both precise and accurte |
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Answer» Correct Answer - B Reading of student A are close to each other and also close to correct reading. Hence reading of A are precise also. |
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| 199. |
Number of milli moles in 1.0 gram of waterA. `1.0`B. 18C. `55.55`D. 100 |
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Answer» Correct Answer - C No of milli moles = W in mg/MW |
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| 200. |
The gas having same number of molecules as 16g of oxygen isA. 16g of `O_(3)`B. 16g of `SO_(3)`C. 48g of `SO_(3)`D. 1 gm of hydrogen |
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Answer» Correct Answer - D No. of molecules `=(W)/(MW) xx N` |
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