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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.
| 101. |
What is the concentration of the solution that results from mixing `40.0` mL of `0.200` M HCl with `60.0` mL of `0.100` M NaOH?(You may assume the volume are additive.)A. `0.150` M NaClB. `0.0200` M NaCl and `0.0200` M HClC. `0.0200` M NaCl and `0.0600` M HClD. 0.0600 N NaCl and 0.0200 M HCl |
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Answer» Correct Answer - D |
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| 102. |
The atomic weight of `Cu` is `63.546`. There are only two naturally occurring isotopes of copper `.^(63)Cu` and `.^(65)Cu`. The natural abundance of the`.^(63)Cu` isotope must be approximately.A. `10%`B. `30%`C. `50%`D. `72.7%` |
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Answer» Correct Answer - D `.^(63)Cu" ".^(65)Cu` `%` abundance `x" "100-x` Avg mass `=(M_(1)x_(1)+M_(2)x_(2))/(x_(1)+x_(2))` `63.546 = (63xx x + 65(100-x))/(100)` `6354.6=63x+6500-65x` `2x = 145.4 rArr x =70%` . |
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| 103. |
How many grams of `Cu(NO_(3))_(2)` would you need to take to get `1.00 g` of copper? `Cu = 63.5, N = 14, O = 16`. |
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Answer» `Mw` of `Cu(NO_(3))_(2) = 63.5 + 2(14 + 16 xx 3) = 187.5 g` `Cu(NO_(3))_(2) rarr Ci` `1 mol` `1 mol` `= 187.5 g` `= 63.5 g` `63.5 g of Cu` is obtained from `187.5 g of Cu(NO_(3))_(2)` `1 g of Cu` is obtained from `= (187.5)/(63.5) = 2.9528 g` |
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| 104. |
One volume of hydrogen combines with sulphur to produce one volume of gas (A). If the vapour density of (A) be 17, what is its molecular formula ? Volumes have been measured under similar conditions of temperature and pressure. |
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Answer» According to the available information : `{:("Hydrogen + S",rarr,"Gas (A)"),("1 vol",,"1 vol"),("n molecules",,"n molecules"),("1 molecule",,"1 molecule"),("2 atoms",,"1 molecule"):}` Let 2 atoms of hydrogen combine with x atoms of the element sulphur. Therefore, the molecular formula of the gas (A) will be `H_(2)S_(x)` Molar mass of the gas `= 2 xx V.D = 2 xx 17 = 34` (given) By comparison , `2 xx 1 + x xx 32 = 34 or x=1` `:.` Molecular formula of gas (A) `= H_(2)S`. |
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| 105. |
The average atomic mass of copper is `63.5`. It exists as two isotopes which are `._(29)^(63)Cu and ._(29)^(65)Cu`. Calculate the percentage of each isotope present in it. |
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Answer» Let the percentage of `._(29)^(63)Cu` isotope `= x` `:.` The percentage of `._(29)^(65) Cu` isotope `= 100 -x` From the above data, the average atomic mass of `Cu=(63xxx)/(100)+(65xx(100-x))/(100)` But the given average atomic mass of `Cu = 63.5` `:. (63xxx)/(100)+(65xx(100-x))/(100)=63.5` `63x+6500-65x=6350" "2x=6350-6500=-150` `2x=150orx=75` `:.` Percentage of `._(29)&(63)Cu` isotope `= 75 %` , Percentage of `._(29)^(65) Cu` isotope `= 25 %`. |
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| 106. |
How many moles of `CH_(4)` will produce 12.o ethane according to the reaction: `CH_(4) + Cl_(2) overset(80%to CH_(3)Cl + HCl` `2CH_(3)Cl + 2Na overset(50%)to CH_(3)CH_(3) + 2NaCl`A. 2B. 0.8C. 0.32D. 1 |
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Answer» Correct Answer - A |
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| 107. |
In one moleof ethane `(C_(2)H_(6))`, calculate the following: a. Number of moles of carbon atoms b. Number of moles of hydrogen stoms c. Number of molecules of ethane. |
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Answer» (i) Gram atoms of `C = 3 xx 2 = 6` (ii) Gram atoms of Hydrogen `=3 xx 6 =18` (iii) Number of molecules of `C_(2)H_(6) = 3N_(A)` . |
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| 108. |
Mole fraction of `I_(2)` in `C_(6) H_(6)` is 0.2. Calculate molality of `I_(2)` in `C_(6) H_(6)`. `(Mw of C_(6) H_(6) = 78 g mol^(-1))` |
| Answer» `m = (x_(2) xx 1000)/(x_(1) xx Mw_(1)) = (0.2 xx 1000)/(0.8 xx 78) = 3.205` | |
| 109. |
Which of the following `is//are` correct? `100 mL` of `3.0 M HClO_(3)` reacts with excess of `Ba(OH)_(2)` according to the equation: `Ba(OH)_(2) + 2 HClO_(3) rarr Ba (ClO_(3)) + 2H_(2) O` `(Mw` of `Ba(ClO_(3))_(2) = 304 g mol^(-1))`A. 1.5 mol of `Ba(ClO_(3))_(2)` is formedB. 3 mol of `Ba(ClO_(3))_(2)` is formedC. `45.6 g` of `Ba(ClO_(3))_(2)` is obtainedD. `4.56 g` of `Ba(ClO_(3))_(2)` is obtained. |
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Answer» Correct Answer - A::C a. mmoles of `HClO_(3) = 100 xx 3 = 300 "mmol"` `= (300)/(1) "mmol of" Ba(ClO_(3))_(2)` `= 150 "mmol" = 1.5 "mol"` Wrong. Weight of `Ba(ClO_(3))_(2) = 150 xx 10^(-3) xx 304 = 45.6 g` Wrong. |
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| 110. |
Calculate the molality of `1 L` solution of `93% H_(2)SO_(4)` (Weight/volume) The density of the solution is `1.84 g`. |
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Answer» Correct Answer - A::B::D Density `= 1.84 g L^(-1)` Mass of `100 mL` solution `= 100 xx 1.84 = 184 g` Mass of `H_(2) SO_(4)` in `100 mL` solution `= 93 g` Mass of water is `100 mL` solution `= 184 - 93 = 91 g` Molality `= (93)/(98) xx (1)/(91) xx 1000 = 10.428` |
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| 111. |
Calculate molality of 1 litre solution of 93% `H_(2) SO_(4)` by volume. The density of solution is `1.84 g mL^(-1)`. |
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Answer» Given `H_(2) SO_(4)` is 93% by volume `:.` Weight of `H_(2) SO_(4) = 93 g` Volume of solution `= 100 mL` Weight of solution `= 100 xx 1.84 = 184 g` Weight of water `= 184 - 93 = 91 g` Molality `(m) = (W_(2) xx 1000)/(Mw_(2) xx W_(1)) = (93 xx 1000)/(98 xx 91) = 10.42` |
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| 112. |
Calculate the molality and molarity of 93 % `H_(2)SO_(4)` (weight/volume). The density of the solution is `1.84 gmL^(-1)` |
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Answer» Step I. Calculation of molarity of the solution Mass of `H_(2)SO_(4)=93` g Molar mass of `H_(2)SO_(4)=2xx1+32 +4 xx 16 = 98 u =(98 "g mol"^(-1))` Volume of solution `= 100 mL = (100)/(1000)=0.1 L` Molarity of the solution `(M)=("Mass of "H_(2)SO_(4)//"Molar mass")/("Volume of solution in litres")=((93g)//(98"g mol"^(-1)))/((0.1 L))` `9.49"mol L"^(-1)=9.49 M` Step II. Calculation of molality of the solution Mass of 100 mL of solution `= d xx V = (1.84 "g mL"^(-1))xx(100 mL)=184 g` Mass of `H_(2)SO_(4)=93 g` Mass of water in the solution `= 184-93 = 91 g = (91)/(1000)=0.091 kg` Molality of the solution `(m)=("No. of moles of "H_(2)SO_(4))/("Mass of water in kg")=((93g)//(98"g mol"^(-1)))/((0.091kg))` `=10.43"mol kg"^(-1)=10.43 m`. |
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| 113. |
20 mL of a gaseous hydrocarbon was exploded with 120 mL of oxygen . A contraction of 60 mL was observed and a further contraction of 60 mL took place when KOH was added. What is the formula of the hydrocarbon?A. `C_(3)H_(6)`B. `C_(3)H_(8)`C. `C_(2)H_(6)`D. `C_(4)H_(10)` |
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Answer» Correct Answer - B |
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| 114. |
What is molality of 1 M solution of sodium nitrate if its density is `1.25 "g cm"^(-3)` ? |
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Answer» Volume of solution `= 1000 cm^(3)` Density of solution `= 1.25 "g cm"^(-3)` Mass of `1000 "cm"^(3)` of solution `= ("1000 cm"^(3))xx("1.25 g cm"^(-3))=1250 g` Mass of `NaNO_(3)` (molar mass ) present `= 23+14 + 48 = 85` g Mass of solvent (water) `= (1250 - 85)=1165 g = 1.165 kg` Molality of solution `(m)=("Mass of "NaNO_(3)//"Molar mass")/("Mass of solvent in kg")` `=((85g)//(85 "g mol"^(-1)))/((1.165 kg))=0.858"mol kg"^(-1)=0.858` m. |
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| 115. |
`50mL` of `CO` is mixed with `20mL` of oxygen and sparked. After the reaction, the mixture is treated with an aqueous `KOH` solution. Choose the correct option :A. the volume of `CO` that reacts `= 30 mL`B. volume of `CO_(2)` formed `=50mL`C. volume of `CO` that remains after treatment with `KOH = 10mL`D. the volume of the `CO` that remains after treatment with `KOH = 20mL` |
| Answer» Correct Answer - C | |
| 116. |
The total number of neutrons present in `10g D_(2)O (D "is" _(1).^(2)H)` are .A. `2.5`B. `5.0`C. `10.0`D. none of these |
| Answer» Correct Answer - D | |
| 117. |
The % by volume of `C_(4)H_(10)` in a gaseous mixture of `C_(4)H_(10)` and `CO` is 40. When `200ml` of the mixture is burnt in excess of `O_(2)`. Find volume (in ml) of `CO_(2)` produced.A. `220`B. `340`C. `440`D. `560` |
| Answer» Correct Answer - C | |
| 118. |
`200ml` of a gaseous mixture containing `CO,CO_(2)` and `N_(2)` on complete combusion in just sufficient amount of `O_(2)` showed contration of `40ml`. When the resulting gases were passed through `KOH` solution it reduces by `50%` then calculate the volume ratio of `V_(co_(2)): V_(CO): V_(N_(2))` in original mixture .A. `4:1:5`B. `2:3:5`C. `1:4:5`D. `1:3:5` |
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Answer» Correct Answer - C `{:(,N_(x)O_(y),+,yH_(2),rarr,(x)/(2)N_(2)(g),+,yH_(2)O(l)),(,10ml,,30ml,,10ml,,):}` `10(x)/(2)=10" "implies" x=2"` `10y=30" "implies" "y=3` `N_(2)O_(3)` |
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| 119. |
If the fertilizers are priced according to their nitrogen content, which will be the least expensive per 50 kg bag ? (a) Urea, `(NH_(2))_(2))CO` (b) Ammonia `(NH_(3))` (c) Ammonium nitrate `NH_(4)NO_(3)` |
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Answer» The fertilizer with least precentage of nitrogen is expected to be least expensive (a) Gram molecular mass of `(NH_(2))_(2)CO=2xx` Atomic mass of N + Atomic mass of O + Atomic mass of `C+ 4xx` Atomic mas of `H = 2xx 14+ 16 + 12 + 4 xx1 = 60` g Percentage of `N=((28g))/((60g))xx100=46.7%` (b) Gram molecular mass of `NH_(3)`= Atomic mass of `N xx 3` Atomic mass of `H = 14 + 3 = 17 g` Percentage of `N=((14g))/((17g))xx100=82.3%` (c) Gram molecular mass of `NH_(4)NO_(3)=2xx` Atomic mass of `N+3xx` Atomic mass `O + 4 xx` Atomic mass of H `= 2xx14 + 3 xx 16+4 xx1 = 80` g Percentage of `N=((28g))/((80g))xx100=35%` Ammonium nitraten with least percentage of N is the least expensive. |
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| 120. |
`10mL` of gaseous hydrocarbon on combustion gives `40ml` of `CO_(2)(g)` and `50mL` of `H_(2)O` (vapour) The hydrocarbon is .A. `C_(4)H_(5)`B. `C_(8)H_(10)`C. `C_(4)H_(8)`D. `C_(4)H_(10)` |
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Answer» `{:(C_(a)H_(b),+,(a+(b)/(4))O_(2),rarr,aCO_(2),+,(b)/(2)H_(2)O,),(10,rarr,"excess",,-,,-,),(0,,"excess",,10a,,5b,):}` Therefore, `10 a = 40` So, `a = 4 " " rArr 5b = 50 rArr b = 10` `:.` (D) |
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| 121. |
A precipitate of `AgCl` and `AgBr` weighs `0.4066g`. On heating in a current of chlorine, the `AgBr` is converted to `AgCl` and the mixutre loses `0.0725 g` in weight. Find the `%` of `Cl` in original mixture. |
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Answer» Let the masses of `AgCl` and `AgBr` in the original sample be a and b respectively, Therefore `a+b = 0.4066 g` On passing `Cl_(2)` gas through the mixture, AgBr gets converted to `AgCl` `underset(2(108+80))(2AgBr+Cl_(2))rarrunderset(2(108+35.5))(2Agcl+Br_(2))` Now, 188 g of AgBr from AgCl `= 143.5g` `:.` bg of AgBr forms `=((143.5g))/((188g))xx(bg)` Total mass of AgCl after the reaction `= (ag)+((143.5g))/((188g))xx(bg)` Actual mass of AgCl after the reacction `=(0.4066 - 0.0725)=0.3341` g `:. a+(143.5)/(88)xxbg=0.3341g` Subtract equation (ii) from equation (i) `b-(143.5)/(188)b=0.4066-0.3341=0.0725g` `b-0.7633b=0.0725g` `0.2367b=0.0725g` `:.` Mass of `AgBr(b)=(0.0725g)/(0.2367)=0.3063` g Mass of `AgCl (a) = 0.4066 - 0.3063 = 0.1003` g `AgCl -= Cl` `143.5 g = 35.5 g` Mass of chlorine (Cl) present `= ((35.5g))/((143.5g))xx(0.1003g)=0.025`g Percentage of chlorine (Cl) in the original sample `= ((0.025g))/((0.4066g))xx100=6.15%`. |
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| 122. |
What is the normarlity and nature of a mixutre obtained by mixing `0.62 g` of `Na_(2) CO_(3). H_(2) O` to `100 mL` of `0.1 N H_(2) SO_(4)`? |
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Answer» mEq of `Na_(2) CO_(3) H_(2) O = (0.62)/(62) xx 1000 ((W)/(E) xx 1000 = mEq)` mEq of `H_(2) SO_(4) = 100 xx 0.1 = 10` `Na_(2)CO_(3) + H_(2)SO_(4) rarr Na_(2) SO_(4) _ H_(2) O + CO_(2) uarr` mEq added 10 10 0 0 0 mEq left 0 0 10 10 10 `:. N_(Na_(2)SO_(4)) = (10)/(100) = 0.1` Solution becomes neutral since both acid and base are used up and `Na_(2) SO_(4)` does not show hydrolysis. |
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| 123. |
The room temperature in celsius scale is `25^(@)C`. Convert it into the other two scales of measurement. |
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Answer» Temperature on Kelvin scale `= 25+273 = 298 K` Temperature of Farrenheit scale `= 9//5 xx 25 + 32 = 77^(@)F`. |
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| 124. |
The body temperature of a normal healthy person is `98.4^(@)F`. What is the temperature on the celsius scale ? |
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Answer» `.^(@)F=9//5 (.^(@)C)+32^(@)` `98.4 = 9//5 (.^(@)C)+32^(@)` ` .^(@)C=(98.4-32)xx5//9=66.4xx5//9=36.89^(@)C` |
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| 125. |
24 gms of carbon reacts with 38.4 gms of oxygen gas such that no reactant remain.Calculate moles of carbon mono-oxide obtained in the reaction ?A. 2 molesB. 1.2 molesC. 2.4 molesD. 1.6 moles |
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Answer» Correct Answer - D |
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| 126. |
9 " mL of " a mixture of methane and ethylene was exploded with 30 mL (excess) of oxygen. After cooling, the volume was 21.0 mL. Further treatment with caustic potash solution reduced the volume to 7.0 mL. Determine the composition of the mixture. |
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Answer» Correct Answer - `(b) 50` `CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O` `a" "2a" "a` `C_(n)H_(2n-2)+((3n-1)/(2))O_(2)rarrnCO_(2)+(n-1)H_(2)O` `(20-a)((3n-1)/(2))(20-a)n(20-a)` For methane `a+n(20-a)=40….(1)` For oxygen `[100-2a-((3n-1)/(2))(20-a)]=40` `2a+((3n-1)/(2))(20-a)=60` `2a+30n-1.5na-10+0.5a=60` `2.5a-1.5na+30n-70` `2.5a-1.5n(a-20)=70` `2.5a+1.5n(20-a)=70 " "...(2)` from `(1)&(2)` `a=10` `n=3` `C_(3)H_(4)` `%` composition `rarr 50%` |
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| 127. |
Mixture of `MgCO_(3)` &`NaHCO_(3)` on strong heating gives `CO_(2)` &`H_(2)O` in 3 : 1 mole ratio. The weight % of `NaHCO_(3)` present in the mixture is :A. 0.3B. 0.8C. 0.4D. 0.5 |
| Answer» Correct Answer - D | |
| 128. |
280 g of a mixture containing `CH_(4) and C_(2)H_(6)` in `5:2` molar ratio is burnt in presence of excess of oxygen. Calculate total moles of `CO_(2)` produced.A. 9B. 18C. 7D. 12 |
| Answer» Correct Answer - B | |
| 129. |
An hourly requirement of an astronaut can be satisfied by the energy released when 34 grams of sucrose `(C_(12)H_(22)O_(11))` are burnt in his body. How many grams of oxygen would be require to carry in space capsule to meet his requirement for one day (24 hours) ? |
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Answer» Correct Answer - 916.2 g Step I. Mass of sucrose required Mass of sucrose needed for 1 hour = 34 g Mass of sucrose needed for 24 hours `= 34 xx 24 = 816 g` Step II. Mass of oxygen required `underset(underset(=342g)(12xx12+22xx1+16xx11))(C_(12)H_(22)O_(11))+underset(underset(=384g)(12xx32))(12O_(2))rarr12CO_(2)+11H_(2)O` 342 g of sucrose require oxygen = 384 g 816 g of sucrose require oxygen `= (384)/(342)xx816=916.2g`. |
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| 130. |
A `10 gram` sample of natural gas containing `CH_(4)` and `C_(2)H_(4)` was burnt in excess of oxygen to give `29.0 "grams"` of `CO_(2)` and some water. How many games of water are formed :A. `9.42 g`B. `18.81 g`C. `11.42 g`D. `15.31 g` |
| Answer» Correct Answer - B | |
| 131. |
Calculate the mass of : (i) 1.2 gram atom of oxygen (ii) 5.2 gram atom of iodine (iii) 5.6 gram atom of chlorine. |
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Answer» Correct Answer - (i) 19.2 g (ii) 660.4 g (iii) 198.8 g (i) 1.0 gram atom of oxygen = 16.0 g 1.2 gram atoms of oxygen `= 16.0 xx 1.2 = 19.2 g` (ii) 1.0 gram atom of iodine = 127.0 g 5.2 gram atom of iodine `= 5.2 xx 127.0 = 660.4 g` (iii) 1.0 gram atom of chlorine `= 35.5 g` 5.6 gram atom of chlorine `= 5.6 xx 35.5 = 198.8 g` |
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| 132. |
A compound on analysis was found to contain the following composition : `Na=14.31%,S=9.97%,O=69.50 %and H=6.22 %` Calculate the molecular formula of the compound assuming that the whole of hydrogen in the compound is present as water of crystallisation. Molecular mass of the compound is 322. |
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Answer» Determination of empirical formula of the compound `{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("Na",14.31,23,(14.31)/(23)=0.622,(0.622)/(0.311)=2,2),("S",9.97,32,(9.97)/(32)=0.311,(0.311)/(0.311)=1,1),("H",6.22,1,(6.22)/(1)=6.22,(6.22)/(0.311)=20,20),("O",69.50,16,(69.50)/(16)=4.34,(4.34)/(0.311)=14,14):}` The empirical of the compound `= Na_(2)SH_(20)O_(14)` Step II. Determination of molecular formula of the compound Empirical formula mass `= 2xx 23+32+20xx1+14xx16` `=46+32+20+224=332` Molecular mass 322 (Given) `:. n=("Molecular mass")/("Empirical formula mass")=(322)/(322)=1` Molecular formula `= n xx` Empirical formula `= 1 xx Na_(2)SH_(20)O_(14)=Na_(2)SH_(20)O_(14)` The whole of hydrogen is present in combination with oxygen as molecules of water of crystallisation `:.` The number of `H_(2)O` molecules = 10 No. of oxygen atoms not involved in the formation of `H_(2)O` molecules = 4 `:.` Molecular formula of the compound `= Na_(2)SO_(4).10 H_(2)O` |
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| 133. |
Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc, Following reaction takes place `Zn+2HClrarrZnCl_(2)+H_(2)` Calculate the voluem of hydrogen gas liberated at STP when 32.65 g of zinc reacts with HCl. 1 mol of a gas occupies 22.7 L volume at STP, atomic mass of Zn=65 .3u |
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Answer» `underset(65.u (65.3g))(Zn(s)+2HCl(aq))overset("S.T.P")(rarr)Zn.Cl_(2)(aq)+underset(22.7L)(H_(2)(g))` 65.3 g of zinc evolve hydrogen gas at S.T.P = 22.7 L 32.65 g of zinc evolve hydrogen gas at S.T.P `= ((32.65g))/((65.3g))xx(22.7L)=11.35L`. |
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| 134. |
The maximum amount of `BaSO_(4)` precipitated on mixing `BaCl_(2)(0.5 M)` with `H_(2)SO_(4) (1M)` will correspond to :A. 0.05 MB. 0.5 MC. 1.0 MD. 2.0 M |
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Answer» Correct Answer - B `underset(("0.5 M"))(BaCl_(2))+underset(("1M"))(H_(2)SO_(4))rarrunderset(("0.5 M"))(BaSO_(4))+2HCl` `BaCl_(2_` is the limiting reactant. |
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| 135. |
Aluminium reacts with sulphure to form a aluminium sulphide . If 5.4 gm of Aluminium reacts with 12.8 gm sulphure gives 12 gm of aluminium sulphides , then the percent yield of the reaction is- |
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Answer» `2Al + 2S to Al_(2)S_(3)` Mole taken `{:((5.4)/(27) gm ,(12.8)/(32)),(=0.2,=0.4):}` `("mole taken")/("stoich coeff"){:((0.2)/(2),(0.4)/(3)),(=0.1,=0.133),((L.R.),):}` moles of `Al_(2)S_(3)` formed `=(1)/(2)xx0.2=0.1` mass=`0.1xx150=15` gm `%` yield `=("actual yield")/("theoritical yield")xx100=(12)/(15)xx100=80%` |
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| 136. |
How many moles of `KMnO_(4)` are needed to oxidise a mixture of 1 mole of each `FeSO_(4) & FeC_(2)O_(4)` in acidic medium `:`A. `(4)/(5)`B. `(5)/(4)`C. `(3)/(4)`D. `(5)/(3)` |
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Answer» Correct Answer - 1 Equivalent of `KMnO_(4)=` equivalent of `FeSO_(4)+` equivalent of `FeC_(2)O_(4)` `x xx5=1xx1+1xx3` `x=(4)/(5) `mole |
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| 137. |
1.0 g of magnesium is burnt with 0.56 g of `O_(2)` in a closed vessel. Which reactant is left in excess and how much ?A. Mg , 0.44 gB. `O_(2)` , 0.28 gC. Mg , 0.16 gD. `O_(2)` , 0.16 g |
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Answer» Correct Answer - C `underset(2xx24g)(2Mg)+underset(32g)(O_(2))rarr2MgO` 32 g of `O_(2)` react with Mg = 48 g 0.56 g of `O_(2)` react with `Mg=((48g))/((32g))xx(0.56g)` Mass of mg taken `= 1.0 g` Mass of Mg left `= (1- 0.84) = 0.16 g`. |
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| 138. |
In an organic compound of molar mass `108 gm mol^-1 C, H and N` atoms are presents in `9 : 1 : 3.5` by mass. Molecular can beA. `C_(6)H_(8)N_(2)`B. `C_(7)H_(10)N`C. `C_(5)H_(6)N_(3)`D. `C_(4)H_(18)N_(3)` |
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Answer» Correct Answer - A `{:(,C,,H,,N),(,9,:,1,:,3.5),("Mole",(9)/(12),:,(1)/(1),:,(3.5)/(14)),(,0.75,:,1,:,0.25):}` `C_(3)H_(4)N=` emp. Formula mol. Formula `=C_(6)H_(6)N_(2)` |
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| 139. |
Calculate number of atoms in each of the following" i. 0.5 mol atom of nitorgen ii. 0.2 mol molecules of hydrogen iii. 3.2 g of sulphur Calculate number of molecules in each of the following: i. 14 g of nitrogen ii. 3.4 g of `H_(2) S` |
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Answer» 1 mol of `N_(2) = 6.023 xx 10^(23)` molecules or mole atom `0.5 " mol atom of" N_(2) = 6.023 xx 10^(23) xx 0.5` `3.10 xx 10^(23)` atoms ii. 1 mol of `H_(2) = 6.023 xx 10^(23)` molecules `= 2 xx 6.023 xx 10^(23)` atoms `0.2 mol of H_(2) = 2 xx 6.023 xx 10^(23)` atoms iii. `32 g of S = 6.023 xx 10^(23)` atoms `3.2 g of S = 6.023 xx 10^(22)` atoms b. `1 mol of N_(2) = 28 g` i. `28 g of N_(2) = 6.023 xx 10^(23)` molecules `14 g of N_(2) = (6.023 xx 10^(23))/(28) xx 14` `= 3.01 xx 10^(23)` molecules ltbRgt ii. `34 g of H_(2) S implies 6.023 xx 10^(23)` molecules `3.4 g of H_(2) Simplies 6.023 xx 10^(22)` molecules |
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| 140. |
What is the mole fraction of the solute in a 1.0 m aqueous solution ?A. `1.770`B. `0.0354`C. `0.0177`D. `0.177` |
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Answer» Correct Answer - C 1.0 m (Molal) solution implies that 1 mole of the solute is dissolved in 1000 g of the solvent (water) Mole fraction of solute `= (("1 mol"))/(1" mol"+((100g))/(("18 g mol"^(-1))))=(1)/(56.5)` `=0.0177`. |
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| 141. |
20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample ? (At. Wt of Mg = 24)A. 96B. 84C. 60D. 75 |
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Answer» Correct Answer - B `underset(84g)(MgCO_(3))overset(heat)(rarr)underset(40g)(MgO)+CO_(2)` 84 g of `MgCO_(3)` yield upon heating MgO = 40 g 20 g of `MgCO_(3)` yield upon heating `MgO=((40g)xx(20g))/((84g))` `=9.52g` Weight of MgO actually formed = 8.0g % purity of `MgCO_(3)=((8.0g))/((9.52g))xx100=84%`. |
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| 142. |
The weight of `2.01xx10^(23)` molecules of `CO` is-A. `9.3 g`B. `7.2g`C. `1.2g`D. `3g` |
| Answer» Correct Answer - A | |
| 143. |
`9 xx 10^(22)` atoms of Ar and n moles of `O_(2)` are kept in a vessel of capacity 5L at 1 atm and `27^(@)C` . (Consider `N_(A)=6 xx 10^(23), R=0.0821 L atm "mol"^(-1)K^(-1)`) : Find the mass of `O_(2)` in vessel:A. 17 gmB. 3.4 gmC. 1.7 gmD. 34 gm |
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Answer» Correct Answer - C |
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| 144. |
One atom of an element `x` weight `6.643xx10^(-23)g`. Number of moles of atom in `20` kg is :A. 4B. 40C. 100D. 500 |
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Answer» Correct Answer - 4 Atomic weight of an element `x=6.643xx10^(-23)xxN_(A)=40` Number of moles of `x=(20xx1000)/(40)=500` |
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| 145. |
Calculate the number of moles in `5.75 g` of sodium. `("Atomic mass of sodium" = 23)` |
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Answer» Correct Answer - `0.25` mol `n_(Na)=(5.75)/(23)=0.25` mole |
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| 146. |
Calculate the percentage composition of potassium hydrogen sulphate `(KHSO_(4))` |
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Answer» Correct Answer - `K = 28.68 %, H=0.735 %, S = 23.53 %, O 47.06 % Molecular mass of `KHSO_(4)=39+1=32+64=136.0u` Percentage of potassium `= ((39u))/((136u))xx100=28.68%` Percentage of hydrogen `= ((1u))/((136u))xx100=0.735%` Percentage of sulphur `= ((32u))/((136u))xx100=23.53%` Percentage of oxygen `= ((64u))/((136u))xx100=47.06%`. |
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| 147. |
The number of `H_(2)O` molecules in a drop of water weighing 0.018 g is :A. `6.022 xx 10^(26)`B. `6.022 xx 10^(23)`C. `6.022 xx 10^(19)`D. `6.022 xx 10^(20)` |
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Answer» Correct Answer - D Molar mass of water `(H_(2)O)=18 g` No. of `H_(2)O` drops present in 18g of water `= 6.022 xx 10^(23)` No. of `H_(2)O` drops present in 0.018 of water `= ((0.018g))/((18.0g))xx6.022xx10^(23)=6.022xx10^(20)`. |
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| 148. |
No. of `H_(2)O` molecules in a drop of water weighing 0.05 g is :A. `1.15 xx 10^(23)`B. `1.672 xx 10^(21)`C. `1.5 xx 10^(20)`D. `6.022 xx 10^(22)` |
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Answer» Correct Answer - B 18.0 g of `H_(2)O` contain molecules `=6.022 xx 10^(23)` 0.05 g of `H_(2)O` contain molecules `= ((0.05g))/((18.0g))xx6.022xx10^(23)` `=1.673xx10^(21)` molecules. |
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| 149. |
Copper sulphate crystals contains `25.45 % Cu and 36.07 % H_(2)O`. If the law of constant composition is true, calculate the weight of copper required to obtained `40 g` of crystalline copper sulphate. |
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Answer» The chemical formula of copper sulphite crystals is `CuSO_(4).5H_(2)O`. According to the data, Percentage of `Cu = 25.45 % ` , Percentage of `H_(2)O = 36.07 %` Percentage of `SO_(4) = 100-(25.45 +36.07) = 100-61.52 = 38.48 %` Since the law of constant composition is true, the percentage composition of the other sample of copper sulphite crystals must also remais the same Now, 100 g of copper sulphate crystals contain `Cu = 25.45 g` `:.` 50 g of copper crystals contain `Cu=((25.45g))/((100g))xx(40g)=10.18g`. |
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| 150. |
Calculate the no. of molecules in a drop of water weighing `0.07 g`. |
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Answer» Correct Answer - `2.34xx10^(21)"molecules of" H_(2)O` molecular `=((0.07)/(18)) xx N_(A) xx 3=2.34 xx 10^(21)` |
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