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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.
| 1101. |
Calculate the mass percent `(w//w)` of sulphuric acid in a solution prepared by dissovles 4 g of sulphur trixoide in a 100ml of sulphuric acid solution containing 80 mass percent `(w//w)` of `H_2SO_4` and having a density of 1.96 g/ml. (Molecular weight of `H_2SO_4=98 gm`) Taken reaction `SO_3+H_2OtoH_2SO_4`A. 0.8085B. 0.84C. 0.4165D. None of these |
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Answer» Correct Answer - A |
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| 1102. |
If 4 g of NaOH dissovles in 36g of `H_(2)O`, calculate the mole fraction of each component in the solution. (specific gravity of solution is `1g mL^(-1)`). |
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Answer» Step-I Calculation of the mole fractions of the components Mass of NaOH = 40g Molar mass of NaOH `= 40 g mol^(-1)` No. of moles of NaOH `= ("Mass")/("Molar mass")=(("4 g"))/(("40 g mol"^(-1)))=0.1` mol Mass of water = 36 g Molar mass of water `= 18 "g mol"^(-1)` No. of moles of water `= ("Mass")/("Molar mass")=(("36 g"))/(("18 g mol"^(-1)))=2.0` mol Mole fraction of NaOH `= (("0.1 mol"))/((0.1+2.0)"mol")=0.048` Mole fraction of `H_(2)O=(("2.0 mol"))/((0.1+2.0)"mol")=0.952`. Step-II Calculation of the molarity of the solution Mass of the solution = Mass of NaOH + Mass of water `= (4g + 36 g)=40g` Volume of the solution `= ("Mass of the solution")/("Specific gravity of the solution")=((40g))/(("1 gm L"^(-1)))=40mL` Molarity of the solution `(M)=("No. of moles of NaOH")/("Volume of solution in litres")` `= ((0.1 "mol"))/((40//100L))=2.5 "mol L"^(-1)=2.5M`. |
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| 1103. |
Calculate the mass fraction and mole fraction of ethyl alcohol and `H_(2) O` in a solution containing `9.2 g` of alcohol in `18.0 g` of `H_(2) O`. |
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Answer» `X_(2) = (W_(2) // Mw_(2))/((W_(1))/(Mw_(1)) + (W_(2))/(Mw_(2)))` `= (9.2 // 46)/((18)/(18) + (9.2)/(46)) = (0.2)/(1 + 02) = 0.166 ~~ 0.17` Mass fraction `X_(2) = (W_(2))/(W_(1) + W_(2)) = (9.2)/(18 + 9.2)` `= 0.338 = 0.34` |
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| 1104. |
81 gm mixture of `mgCO_(3)(s)` and `NH_(3)CO_(3)(s)` is heated to constant mass. If vapour density of gaseous mixture evolved was found to be `(61)/(4)` then Given that: `MgCo_(3)(s) rarr MgO_(s) + CO_(2)(g)` `NH_(2)COONH_(4)(s) rarr 2NH_(3)(g)+CO_(2)(g)` Mole % of `MgCo_(3)` in original sample:A. 0.5B. 0.6C. 0.75D. None of these |
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Answer» Correct Answer - A |
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| 1105. |
A 96 gm mixture containing `CaCO_(3) "and" MgCO_(3)` on heating produces a gas which dissolves in 1 litre of water to form 1 `M H_(2)CO_(3)` solution. Mole ratio of `CaCO_(3) "and" MgCO_(3)` in original mixture is: |
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Answer» Correct Answer - 3 |
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| 1106. |
How many `mL` of a `0.1M HCl` are required to react completely with `1 g` mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` containing equimolar amounts of two? |
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Answer» Step I. Calculation of mass of constituents in the mixture Mass of mixture = 1.0 g Let the mass of `Na_(2)CO_(3) = xg` Mass of `NaHCO_(3)=(1-x)g` Moles of `Na_(2)CO_(3)=("Mass of "Na_(2)CO_(3))/("Molar mass")=((xg))/(("106 g mol"^(-1)))=(x)/(106)` mol Moles of `NaHCO_(3)=("Mass of "NaHCO_(3))/("Molar mass")=((1-x)g)/(("84 g mol"^(-1)))=(1-x)/(84)`mol According to available data : Moles of `Na_(2)CO_(3)` = Moles of `NaHCO_(3)` `:. ((x)/(106)"mol")=((1-x)/(84)"mol")` or `84x=106-106xorx=(106)/(190)=0.558g` `:.` Mass of `Na_(2)CO_(3)` in the mixture = 0.558 g Mass of `NaHCO_(3)` in the mixture `= (1 - 0.558) =0.442 g` Step II. Calculation of total mass of `HCl` required `underset(106g)(Na_(2)CO_(3)(s))+underset(73g)(2HCl)(aq)rarr2NaCl(aq)+H_(2)O(l)+CO_(2)(g)` `underset(84g)(NaHCO_(3))(s)+underset(36.5g)(HCl(aq))rarrNaCl(aq)+H_(2)O(l)+CO_(2)(g)` Now, 106 g of `Na_(2)CO_(3)` require HCl = 73 g 0.558 g of `Na_(2)CO_(3)` require `HCl=((73g)xx(0.558g))/((106g))=0.384g` Similarly, 84 g of `NaHCO_(3)` require HCl = 36.5 0.442 g of `NaHCO_(3)` require HCl `= ((36.5g)xx(0.442g))/((84g))=0.192g` `:.` Total mass of HCl required `= (0.384 + 0.192) = 0.576 g` Step III. Calculation of volume of `HCl` required Mass of HCl required = 0.576 g Molarity of HCl solution = 0.1 M Molarity of solution (M) `= ("Mass of HCl/Molar mass of HCl")/("Volume of HCl solution (V)")` `("0.1 mol L"^(-1))=((0.576g)//(36.5g//mol))/(V)` `= V = ((0.576g))/(("36.5 gmol"^(-1))xx("0.1 mol L"^(-1)))=0.1578L =157.8mL`. |
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| 1107. |
Calculate the mass of aluminium oxide which contains double the number of oxygen atoms in 192 g of oxygen gas. |
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Answer» (i) atomicity of oxygen (ii) chemical composition of `Al_(2)O_(3)` (iii) determination of mass of `Al_(2)O_(3)` (iv) 408 g |
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| 1108. |
The percentage of oxygen in a metallic oxide of a bivalent metal is `20.1%`. The molecular weight of the compound is `79.5`. Write the molecular formular of the compound, considering the symbol of the metal as M and find out the atomic weight of the metal. |
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Answer» (i) calculate the number of atoms of oxygen present in the compound (ii) calculation of the number of atoms of oxygen present in one molecule of the metallic oxide from the percentage of oxygen given (iii) determination of the formula of the compound (iv) calculation of atomic weight of the metal (v) `63.5` |
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| 1109. |
A sample of urine containing `0.3 g` of urea was treated with an excess of `0.2 M` nitrous acid, according to the equation. `NH_(2)CONH_(2) + 2HNO_(2) rarr CO_(2) + 2N_(2) + 2H_(2)O` The gass produced passed through aqueous `KOH` solution and the final valume is measured. (Given, `Mw_("urea") = 60 g mol^(-1)`, molar volume of gas at standard condition, i.e., at room temperature `25^(@)C` and 1 atm pressure. `RTP` (room temperature pressure) also is `24.4 L` or `24400 mL mol^(-1))` What is the volume at `RTP`?A. `122 mL`B. `244 mL`C. `366 mL`D. `488 mL` |
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Answer» Correct Answer - B `NH_(2)CONH_(2) + 2HNO_(2) rarr CO_(2) uarr + 2N_(2) uarr + 3H_(2)O(l)` mmoles `= (0.3 xx 10^(3))/(60)` `= 5` Initial 5 10 - - - Final - - 5 10 15 `V_(H_(2)O)` at room temperature = 0 (since it is liquid). `V_(CO_(2))` is absorbed by `KOH` or `NaOH`. `:.` millimoles of gases (i.e., `N_(2)`) = 10 `V` at `RTP = 24400 xx 10 xx 10^(-3) mL = 244 mL` |
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| 1110. |
An aqueous solution is made by dissolving glucose `(C_(6)H_(12)O_(6))` and urea `(NH_(2)CONH_(2))` in water. Mole ratio of glucose and water is 1:10 . If the masses of glucose and urea are in 3:1 ratio, then correct statement(s) regarding the solution is/are:A. The mole fraction of glucose in the solution is `(1)/(11)`B. The mole fraction of urea in the solution is `(1)/(12)`C. Molality of glucose in the solution is `(25)/(6)`D. Molality of urea in the solution is `(50)/(9)` |
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Answer» Correct Answer - B::D |
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| 1111. |
A 100 mL portion of `0.250` M calcium nitrate solution. What is the final concentration of the nitrate ion?A. `0.180` MB. `0.130` MC. `0.0800` MD. `0.0500` M |
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Answer» Correct Answer - A |
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| 1112. |
Two glucose solution are mixed. One has a volume of `480 mL` and a concentration of `1.50 M` and the second has a volume of `250 mL` and concentration `1.20 M`. The molarity of final solution isA. `2.70` M`B. `1.40` MC. `1.50` MD. `1.20` M |
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Answer» Correct Answer - D |
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| 1113. |
If 87g of `K_(2)SO_(4)` (molar mass=174g) is dissolved in enough water to make 250 mL of solution, calculate sum of concentration of `[K^(+)]+[SO_(4)^(-2).]` |
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Answer» Correct Answer - 6 |
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| 1114. |
5 g of `K_(2)SO_(4)` was dissolved in 250 ml of solution. How many ml of this solution sholuld be used so that 2.33 g of `BaSO_(4)` may be precipitated from `BaCl_(2)` solution. `K_(2)SO_(4)+BaCl_(2)rarrBaSO_(4)+2KCl` |
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Answer» Correct Answer - 87 |
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| 1115. |
The volume of water that must be added to a mixture of 250ml of 0.6 M HCl and 750 ml of 0.2 M HCl to obtain 0.25 M solution of HCl is:A. 750 mlB. 100 mlC. 200 mlD. 300 ml |
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Answer» Correct Answer - C |
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| 1116. |
Calculate the volume of `1.00 mol L^(-1)` aqueous sodium hydroxide that is neutralized by `200 mL` of `2.00 mol L^(-1)` aqueous hydrochloric acid and the mass of sodium chloride produced. Neutralization reaction is, `NaOH_((aq.))+HCl_((aq.))rarr NaCl_((aq.))+H_(2)O_((l))` |
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Answer» Step I. Calculation of volume of NaOH solution neutralised Volume of NaOH solution neutralised can be calculated by applying molarity equation `M_(1)V_(1)=M_(2)V_(2)` `(1 M)xxV_(1) = (2M)xx(200 mL)=400 mL` Step II. Calculation of mass of `NaCl` produced The neutralised reaction is : `NaOH(aq)+underset(underset(36.5g)(("1 mol")))(HCl(aq))rarrunderset(underset(58.5g)(("1 mol")))(NaCl(aq))+H_(2)O(l)` `:.` Mass of HCl in 200 ml of 2M solution can be calculated as follows : molarity of solution `= ("Mass of HCl/Molar mass")/("Volume of solution in litres")` `(2.0"mol L"^(-1))=("Mass of HCl")/(("36.5 g mol"^(-1))xx(0.2L))` Mass of HCl `= (2.0 "mol L"^(-1))xx("36.5 g mol"^(-1))xx(0.2 L)=14.6 g` Now, `36.5 g` of HCl produce NaCl after neutralisation = 58.5 g `:.` 14.6 of HCl produce NaCl after neutralisation `= ((58.5g))/((36.5g))xx(14.6g)=23.4g`. |
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| 1117. |
How much of NaOH is required to neutralise 1500 `cm^(3)` of `0.1` M HCl?A. 40 gB. 4 gC. 6 gD. 60 g |
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Answer» Correct Answer - C |
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| 1118. |
`Fe_(2)O_(3)` reacts with excess CO at a high temperature according to the equation below: `Fe_(2)O_(3) + 3CO rightarrow 2Fe + 3CO_(2)` If 6.5-0g of `Fe_(2)O_(3)` yields 3.85g of Fe, what is the percentage yield of the reaction ?A. 0.592B. 0.699C. 0.763D. 0.847 |
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Answer» Correct Answer - D |
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| 1119. |
25 mL of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a titre value of 35 mL. The molarity of barium hydroxide solution was :A. 0.28B. 0.35C. 0.07D. 0.1 |
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Answer» Correct Answer - C `underset("1 mol")(Ba(OH)_(2))+underset("2 mol")(2HCl)rarrBaCl_(2)+2H_(2)O` No. of moles of `HCl=(("0.1 mol"^(-1)))/(("1L"))xx("0.035L")` `=3.5xx10^(-3)` mol No. of moles of `Ba(OH)_(2)=(3.5xx10^(-3))/(2)` `=1.75xx10^(-3)` mol Molarity of `Ba(OH)_(2)` solution `= ("No. of moles of Ba"("OH")_(2))/("Volume of solution in litres")` `=((1.75xx10^(-3)"mol"))/(("0.025 L"))=0.07"mol L"^(-1)` `=0.07M`. |
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| 1120. |
28 gm lithium is mixed with 48 gm `O_(2)` to reacts according to the following reaction. `4Li + O_(2) to 2Li_(2)O` The mass of `Li_(2)O` formed is |
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Answer» `4Li + O_(2) to 2Li_(2)O` moles taken `{:((28)/(7),(48)/(32)),(=4,=1.5):}` `("moles taken")/("stoich. Coeff") (4)/(4)=1" "(1.5)/(1)=1.5` `" "` (L.R.) Moles of `Li_(2)O` formed =`(2)/(4)x4=2` Mass of `Li_(2)O` formed=`2xx30=60` gm |
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| 1121. |
What mass of `Na_(2)SO_(4).7H_(2)O` contains exactly `6.023xx10^(22)` atoms of oxygen? |
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Answer» Molar mass of `Na_(2)SO_(4).7H_(2)O` =275 gm 1 mole `Na_(2)SO_(4).7H_(2)O` has 11 mol O-atoms. `implies 11 N_(A)O`- atoms are in 275 g `Na_(2)SO_(4).7H_(2)O` `implies 6.023xx10^(23)O`- atoms are in `=(275)/(11xx6.023xx10^(23))xx6.023xx10^(22)` g=2.5 g |
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| 1122. |
Calculate the weight of `CaO` required to remove hardness of `10^(6) L` of water containing `1.62 g` of `Ca (HCO_(3))_(2)` in `1.0 L`. `(Mw of Ca (HCO_(3))_(2) = 162, mw of CaO = 56)` |
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Answer» The reaction is: `underset(1 mol)(CaO) + underset(1 mol)(Ca (HCO_(3))_(2) rarr underset(2 mol)(2CaCO_(3)) + H_(2) O` Moles of `Ca(HCO_(3))_(2)` in `1.0 L` of sample `= (1.62)/(162) = 0.01 mol` Moles of `CaO` required in `1.0 L` of sample `= 0.1 mol L^(-1)` Mole of `CaO` requried in `10^(6) L` of sample `= 0.01 xx 10^(6) mol//10^(6) L = 10^(4) mol (10^(6) L)^(-1)` Weight of `CaO = 10^(4) xx 56 = 5.6 xx 10^(5) g` |
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| 1123. |
How many mL of `8.00` M HCl are needed to prepare 150 mL of a `1.60` M HCl solution?A. `30.0` mLB. `24.0` mLC. `18.8` mLD. `12.0` mL |
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Answer» Correct Answer - A |
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| 1124. |
`0.5` mole of `H_(2)SO_(4)` is mixed with `0.2` mole of `Ca(OH)_(2)`. The maximum number of mole of `CaSO_(4)` formed is:A. 0.2B. 0.5C. 0.4D. 1.5 |
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Answer» Correct Answer - A |
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| 1125. |
When 1 L of `0.1` M sulphuric acid solution is allowed to react with 1 L of `0.1` M sodium hydroxide solution, the amount of sodium sulphate (anhydrous) that can be obtained from the solution fromed and the concentration of `H^(+)` in the solution respectively are :A. `3.55g,0.1M`B. `7.10g,0.025M`C. `3.55g,0.025M`D. `7.10g,0.05M` |
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Answer» Correct Answer - D |
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| 1126. |
In which of the following mixtures summation of molarity of cation(s) in the resulting solution is less than 1?A. `NHO_(3)(0.1 M,10 mL) + NaOH (1 M, 10 mL)`B. `NHO_(3)` (0.1 M, 10mL)+`(NHO_(3)` (1.1 M, 10 mL)C. NaOH (4%(w/w), 10 mL sp.gr.=1.2) + NaOH [4% (w/w) , 10mL, sp.gr.=12]D. `AgNO_(3)` (1M, 10 mL) + NaCl(2M, 10 mL) |
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Answer» Correct Answer - A::B |
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| 1127. |
Which fo the following is heavier than 1 gm molecules oxygen?A. 12 gm of `O_(3)`B. 1 gm-molecules `O_(3)`C. 4 gm-atom of hydrogenD. 1.12 litre of `H_(2)O` at `40^(@)` C and 1 atm |
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Answer» Correct Answer - B::D |
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| 1128. |
Calculate the weight of iron which will be converted into its oxide `(Fe_(3)O_(4))` by the action 18 g of steam on it. |
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Answer» Correct Answer - 42 g Chemical equation for the reaction is : `underset(underset("= 168 g = 73 g")(3xx56" " 4 xx 18))(3Fe+4H_(2)O(g))rarrFe_(3)O_(4)(s)+4H_(2)(g)` 72 g of steam `(H_(2)O)` react with iron = 168 g 18 g of steam `(H_(2)O)` react with iron `= (168xx18)/(72)=42g`. |
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| 1129. |
In one process of waterproofing, a fabric is expsoed to `(CH_(3))_(2) SiCl` vapour. The vapour reacts with `(OH)` groups on the surface of the fabric or with traces of `H_(2) O` to form waterproofing film of by the reaction Where `n` is large integer. The waterproofing film is deposited on the fabric layer upon layer. Each layer is `10 Å` thick [the thickness of the `(CH_(3))_(2) SiO` group]. How much `(CH_(3))_(2) SiCl_(2)` is required to waterproof one side of a piece of a fabric, `1.0 m` by `3.0 m`, with a film 1000 layers thick? The density of the film is `1.0 g cm^(-3)`. (Atomic weight of `Si = 28` and `Cl = 35.5`) |
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Answer» Correct Answer - `0.9413` gram Volume `1xx3xx300xx6xx10^(-10)` `=5.4xx10^(-7)m^(3)=0.54cm^(3)` `p=1g//cm^(3)` mass `=0.54g` `n(CH_(3))_(2)SiCl_(2)+2nOH^(-)rarr2nCl^(-)+nH_(2)O+[(CH_(3))_(2)SiO]_(n)` `(w)/(129)" "(w)/(129n)xx(74n)` `(74w)/(129)=0.54implies w=0.9413 g` |
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| 1130. |
A metal forms two oxides. The higher oxide contains 80% metal. `0.72 g` of the lower oxide gave `0.8 g` of higher oxide when oxidised. Calculate the weight of oxygen the combines with the fixed weight of metal in the two oxides, and show that the data supports the law of multiple proportines |
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Answer» First method: Since, `0.8 g` of higher oxide is obtained from `0.72 g` of lower oxide, therefore, the weight of lower oxide that would produce `100 g` higher oxide on oxidation `= (0.72 xx 100)/(0.80) = 90 g` Thus, `90 g` of lower oxide contains as much metal as `100 g` of higher oxide, i.e., `80 g` (given). Hence, `80 g` of metal combines with `10 g` of oxygen in the lower oxide and `20 g` of oxygen in the higher oxide. The weights of oxygen that combine with the same weight of the metal in the two oxides are in the ratio of `10 : 20` or `1 : 2`, The ratio, being simple, proves the law of multiple proportions. Second method : Lower oixde `+ O_(2) rarr` Higher oxide Lower oxide : `0.72 g` Weight of oxide `= 0.72 g` Weight of `M` (fixed weight) `= 0.64 g` Weight of `O_(2) = 0.72 - 0.64 = 0.08 g` of `O_(2)` Higher oxide : `0.2 g` `100 g` of oxide contains `implies 80 g of M` `0.8 g` of oxide contains `implies (80)/(100) xx 0.8 implies 0.64 g of M` weight of `O_(2) = 0.8 - 0.64 = 0.16 g of O_(2)` Ratio of oxygen lower and higher oxide `= 0.08 : 0.16 implies 1 : 2` |
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| 1131. |
What is the molecular mass of a compound `X`, if its `3.0115 xx 10^(9)` molucules weigh `1.0 xx 10^(-12) g?` |
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Answer» `3.0115 xx 10^(9)` molecules of `X = 10^(-12) g` `6.023 xx 10^(23)` molecules of `X = (10^(-12) xx 6.023 xx 10^(23))/(3.0115 xx 10^(9))` `200g = 200 amu` |
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| 1132. |
What is the volume occupied by one `C Cl_(4)` molecule at `20^(@)C`? Density of `C Cl_(4)` is `1.6//cc` at `20^(@)C` |
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Answer» `C Cl_(4)` molecule is liquid. `Mw` of `C Cl_(4) = 12 + 4 xx 35.5 = 154 g` 1 molecule of `C Cl_(4) = (154)/(6.023 xx 10^(23)) = 25.56 xx 10^(-23) g` Volume of one molecule `= ("Mass")/("density")` `= (25.56 xx 10^(-23))/(1.6)` `= 1.598 xx 10^(-22) mL or cm^(3)` |
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| 1133. |
10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in the reaction will be :A. 3 molB. 4 molC. 1 molD. 2 mol |
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Answer» Correct Answer - B `underset("1 mol")(H_(2)(g))+underset("0.5 mol")(1//2O_(2)(g))rarrunderset("1 mol")(H_(2)O(l))` 10 gof `H_(2)=((10g))/(("2 g mol"^(-1)))=5` mol 64 g of `O_(2)=((64g))/(("32 g mol"^(-1)))=2` mol In this reaction, `O_(2)` is limiting reactant 0.5 mol of oxygen from water = 1 mol 2 mol of oxygen from water `= (("1 mol"))/(("0.5 mol"))xx("2 mol")=4` mol. |
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| 1134. |
Calcualte the number of oxygen atoms and its weight in 50 gm of `CaCO_(3)`A. `6.02 xx 10^(23)` and 12gmB. `9.033 xx 10^(23)` and 24gmC. `9.033 xx 10^(23)` and 12 gmD. `9.033 xx 10^(23)` and 12gm |
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Answer» Correct Answer - B |
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| 1135. |
10 moles of `A_(2)B_(3)` contains 100 gm A atom and 60 gm B atoms. Choose the correct statements.A. Molecules weight of `A_(2)B_(3)` is equal to 16B. Atomic weight of A is equal to 16C. Weight of one atom of B is equal to 2D. Atomic weight of B is equal to 6 |
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Answer» Correct Answer - A::B |
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| 1136. |
In the reaction `4A + 2B + 3C rArr A_(4)B_(2)C_(3)` what will be the number of moles of product formed ? Starting from 2 moles of A, 1.2 moles of B and 1.44 moles of C.A. 0.5B. 0.6C. 0.48D. 4.64 |
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Answer» Correct Answer - C |
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| 1137. |
A fresh `H_(2)O_(2)` solution is labelled `11.2V`. This solution has the same concentration as a solution which isA. `3.4%` (w/w)B. `3.4%`(v/v)C. `3.4%` (w/v)D. none of these |
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Answer» Correct Answer - C |
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| 1138. |
12 moles of each A and B are allowed to react as given : `3A + 2B rArr C + (1)/(2)D`. If 60g of D is produced then calculate the atomic mass of D.A. 30B. 45C. 60D. 15 |
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Answer» Correct Answer - A |
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| 1139. |
Bottle (A) contains 320 mL of `H_(2)O_(2)` solution and labelled with 10 V `H_(2)O_(2)` and bottle (B) contains 80 mL `H_(2)O_(2)` having molarity 2.5 M. Content of bottle (A) and bottle (B) are mixed and solution is filled in bottle (C). Select the correct label for bottle (C) in terms of volume strength and g/litre. (Assume 1 mole of an ideal gas occupies 22.4 L at STP)A. 13.6 V and `41.276` g/LB. 11.2 V and 0.68 g/LC. 5.6 V and 0.68 g/LD. 5.6 V and 41.286 g/L |
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Answer» Correct Answer - A |
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| 1140. |
A bottle of `H_(3)PO_(4)` solution contains `70%(w//w)` acid. If the density of the `H_(3)PO_(4)` solution required to prepare 1 L of 1N solution is :A. 90 mLB. 45 mLC. 30 mLD. 23 mL |
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Answer» Correct Answer - C |
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| 1141. |
How many `g "of" KCl` would have to be dissolved in `60 g H_(2)O` to give `20%` by weight of solution?A. 15 gB. `1.5` gC. `11.5 g`D. `31.5` g |
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Answer» Correct Answer - A |
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| 1142. |
Equal weight of `NaCl` and `KCl` are dissolved separately in equal of solutions. Molarity of the two solutions will be:A. equalB. that of NaCl will be less than that of KCl solution.C. that of NaCl will be more than that of KCl solutionD. that of NaCl will be half than that of KCI solution |
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Answer» Correct Answer - C |
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| 1143. |
Calculate the molecular mass of the following : (i) `H_(2)O` (ii) `CO_(2)` (iii) `CH_(4)` |
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Answer» Step I. Calculation of mass of methanol `(CH_(3)OH)` Molar mass of methanol `(CH_(3)OH)=12 + 4 xx 1 + 16 = 32 "g mol"^(-1)` Molarity of solution `= 0.25 M = 0.25 "mol L"^(-1)` Volume of solution = 2.5 L Molarity of solution `= ("Mass of methanol/Molar mass")/("Volume of solution in Litres")` `(0.25 "mol L"^(-1))=(W)/((32"g mol"^(-1))xx("2.5 L"))` `W = (0.25 "mol L"^(-1))xx("32 g mol"^(-1))xx("2.5 L")=20 g` Step II. Calculation of volume of methanol Mass of methanol = 20 g = 0.002 kg Density of methanol `= 0.793 "kg L"^(-1)` Volume of methanol `= ("Mass")/("Density")=((0.002kg))/((0.793 "kg L"^(-1)))=0.0025 L`. |
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| 1144. |
Cortisone is a molecular substance substance containing `21` atoms of carbon per molecule. The mas percentage of carbon in cortisone is `69.98%`. Its molar mass is :A. `176.5`B. `252.2`C. `287.6`D. `360.1` |
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Answer» Correct Answer - D `wt` of carbon `=21xx12 g` `because 69.98 g` carbon contain `100 g` cortisone `therefore 1 g` carbon contain `100 g `cortisone `=(100)/(69.98)` `therefore 21 xx 12 g rarr =(100)/(69.98)xx21xx12=360.10` |
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| 1145. |
A compounds contains 69.5% oxygen and 30.5 % nitrogen and its molecular weight is 92. The formula of that compound is :-A. `N_(2)O`B. `NO_(2)`C. `N_(2)O_(4)`D. `N_(2)O_(5)` |
| Answer» Correct Answer - C | |
| 1146. |
lt. Of a hydrocarbon weighs as much as one litre of `CO_(2)` .The molecular formula of the hydrocarbon is :-A. `C_(3)H_(8)`B. `C_(2)H_(6)`C. `C_(2)H_(4)`D. `C_(3)H_(6)` |
| Answer» Correct Answer - A | |
| 1147. |
Which of the following contains the same number of molecules ?A. `1g` of `O_(2)`,`2g` of `SO_(2)`B. `1g` of `CO_(2)` , `1g` of `N_(2)O`C. `112ml` of `O_(2)` at `STP,` `224 ml` of `He` at `0.5atm` and `273K`D. All of these |
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Answer» Correct Answer - 4 `(1)` No. of molecules of `O_(2)=(1)/(32)N_(a)` No. of molecules of `SO_(2)=(2)/(64)Na` or `(1)/(32)N_(a)` `(2)` No. of molecules of `CO_(2)=(1)/(44)N_(a)` No. of molecules of `N_(2)O=(1)/(44)N_(a)` `(3)` No. of molecules of `O_(2)` at `STP=(112)/(22400)=0.0050` No. of molecules of `He=(PV)/(RT)=((224)/(1000)xx0.5)/((1)/(2)xx273)=0.0050` |
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| 1148. |
A sample of oleum is such that ratio of free `SO_(3)` by combined `SO_(3)` is equal to unity. Calculate its labelling in terms of percentage oleum. |
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Answer» Correct Answer - `110.11%` Let free `SO_(3)rarr xg` `SO_(3)` in form of `H_(2)SO_(4)` `rarr(x)/(80)xx98=1.225x` so total `x+1.225x=100` `x=449.49` water required `=(44.94)/(80) xx18 = 10..11 g%` oleum `=100+10.11=110.11%` |
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| 1149. |
One litre of milk weighs. `1.035 kg`. The butter fat is `4%(v//v)` of milk has density of `875 kg//m^(3)`. Find the density of fat free skimed milk. |
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Answer» Correct Answer - `1.041g//mL` `100mL` milk `rarr 4mL` fat `1L` milk `rarr 40mL` fat density of fat `=875kg//m^(3)=0.875g//mL` mass of fat `=40xx0.875=35g` fat free milk mass `=1035-35=1000g` `vol. = 1000-40=960mL` `p=(1000)/(960)=1.0416g//mL` |
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| 1150. |
When `4.215 g` of metallic carbonate was heated in a hard glass tube, the `CO_(2)` evolved was found to be measured `1336 mL` at `27^(@)C` and `700 mm` pressure. What is the equivalent weight of the metal? |
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Answer» Correct Answer - A::B `Pv = nRT = (W)/(Mw) RT` `:. (700)/(760) xx (1336)/(1000) = (W)/(44) xx 0.0821 xx 300` or `W = 2.198 g CO_(2)` The molecular mass of any metal carbonate is `2E + 60`, Where `E` is the equivalent mass of the metal. So `(2E + 60)/(4.215) = (44)/(2.193)` `E = 12.188 g` |
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