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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.
| 651. |
(a) How many gram molecules are present in 4.9 g of `H_(2)SO_(4)` ? (b) Calculate the mass of `0.72` gram molecules of `CO_(2)`. |
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Answer» (a) Molecular mass of `H_(2)SO_(4)= 2xx` Atomic mass of H + Atomic mass of `S + 4 xx` Atomic mass of O `= 2xx1+32+4xx16=98u` Gram molecular mass of `H_(2)SO_(4)=98` g 98g of `H_(2)SO_(4)=1` gram molecule 4.9 g of `H_(2)SO_(4)=(1)/(98)xx4.9 = 0.05` gram molecule (b) Molecular mass of `CO_(2)=` Atomic mass of `C+2xx` Atomic mass of O `= 12+12 xx16=44 u` Gram molecular mass of `CO_(2)=44g` 1 gram molecule of `CO_(2)` has mass `= 44 g` 0.72 gram molecule of `CO_(2)` has mass `= (1)/(44)xx0.72=0.016g`. |
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| 652. |
How many surcose molecules `(C_(12)H_(22)O_(11))` are present in `3.42g` sucrose ? . |
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Answer» Molar mass of sucrose = 342 Molar of sucrose `= (3.42)/(342) = 0.01` Mobeoules of sucrose `= 0.01 N_(A) = 6.023 xx 10^(21)` . |
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| 653. |
Calculate mass of water present in `499gm CuSO_(4).5H_(2)O` ? (Atomic mass `-Cu = 63.5, S = 32,O = 16, H =1)` . |
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Answer» Molar mas of `CuSO_(4).5H_(2)O = 249.5` moles of `CuSO_(4).5H_(2)O = (499)/(249.5) = 2` Each mole `CuSO_(4).5H_(2)O` has `5mol H_(2)O` `2` mole `CuSO_(4).5H_(2)O` has `10 mol H_(2)O` Mass of `10mol H_(2)O = 10 xx 18 = 180 gm` . |
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| 654. |
Calculate the number of atoms in : (i) 0.25 moles atoms of carbon (ii) 0.20 mole molecules of oxygen. |
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Answer» (i) 1 mole atoms of carbon `= 6.022 xx 10^(23)` atoms `:. 0.25` mole atoms of carbon `= 6.022 xx 10^(23) xx 0.25 = 1.506 xx 10^(23)` atoms (ii) 1 mole molecules of oxygen `= 6.022 xx 10^(23)` molecules 0.20 mole molecules of oxygen `= 6.022 xx 10^(23) xx 0.20 = 1.205 xx 10^(23)` molecules. |
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| 655. |
How many atoms of oxygen and hydrogen are present in 0.15 mole of water ? |
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Answer» `underset(("1 mol"))(H_(2)O)underset(("2 gram atoms"))(-=2H)+underset(("1 gram atom"))(O)` (i) No. of hydrogen atoms 1 mole of water `(H_(2)O)` has hydrogen (H) = 2 gram atoms 0.15 mole of water `(H_(2)O)` has hydrogen `(H)=((0.15"mol"))/(("1 mol"))xx("2 gram atoms")=0.30` gram atom No. of hydrogen atoms present `= 0.30 xx 6.022 xx 10^(23) = 1.81 xx 10^(23)` atoms (ii) No. of oxygen atoms 1 mole of water `(H_(2)O)` has oxygen (O) = 1 gram atom 0.15 mole of water `(H_(2)O)` has oxygen `(O) = (("0.15 mol"))/(("1 mol"))xx("1 gram atom")=0.15` gram atom No. of oxygen atoms present `= 0.15 xx6.022xx10^(23)=9.03xx10^(22)` atoms. |
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| 656. |
Find no protons in `180mol H_(2)O` Density of water `= 1 gm//ml` . |
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Answer» Mass of water =density xvolume `= 180g` Moles of water `= (180)/(18) =10` `1` mol water has `10` mol protons `10` mol water has `100` mol protons `10` mol water has `100N_(A)` protons `10` mol water has `6.023 xx 10^(25)` protons . |
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| 657. |
Calculate the no. of moles of phosphorus in 92.9 g phosphorusm assuming that molecular formula of phosphorus is `P_(4)`. Also calculate the no. of atoms and molecules of phosphorus in the sample. |
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Answer» (i) No. of moles of phosphorus `(P_(4))` Molar of phosphorus `(P_(4))=4xx31 = 124 g` Mass of phosphorus `(P_(4))=92.9 g` `:.` No. of moles of `P_(4)=("Mass of "P_(4))/("Molar mass of" P_(4))=((92.9g))/((124"g mol"^(-1)))=0.75` mol (ii) No. of molecules of phosphorous `(P_(4))` 1 mole of `P_(4)` contains molecules `= 6.022 xx 10^(23)` 0.75 mole of `P_(4)` contains molecule `= 6.022 xx 10^(23)xx0.75xx0.75=4,52xx10^(23)` molecules (iii) No. of atoms of phosphorus 1 molecule of `P_(4)` contains atoms = 4 `:. 4.52 xx 10^(23)` molecules of `P_(4)` contain atoms `= 4 xx 4.52 xx 10^(23) =1.8 xx 10^(24)` atoms. |
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| 658. |
Which of the following has the maximum mass (a) 20 g of phosphorus (b) 5 moles of water (c) `12 xx 10^(24)` atoms of hydrogen ? |
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Answer» Correct Answer - 5 moles of water (a) 20 gram of phosphorus = 20.0 g (b) 1 gram mole of water = 18.0 g 5 gram moles moles water `= 18.0 xx 5 = 90.0 g` (c) `6.022 xx 10^(23)` atoms of hydrogen = 1.0 g `12.0 xx 10^(24)` atoms of hydrogen `= (12.0 xx10^(24))/(6.022xx10^(23))=19.93g` 5 moles of water have maximum mass = 90.0 g. |
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| 659. |
Choose the incorrect statement(s).A. 1 gm molecules always contains same number of atomsB. Weight of one molecules in gm is equal to its molar massC. Number of atoms in 2 gm of hydroden is greater than 11.35 litre hydegen at STPD. Volume of 16 gm oxygen gas at 2 atm , 300 K is greater than volume of 2 gm hydrogen gas at 1 atm 300 K |
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Answer» Correct Answer - A::B::D |
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| 660. |
Calculate the mass of the following : (i) 1 atom of carbon (ii) 1 atom of silver (iii) 1 molecule 1 benzene `(C_(6)H_(6))` (iv) 1 molecule of water `(H_(2)O)` |
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Answer» Correct Answer - (i) `1.99 xx 10^(-23)g` (ii) `1.79 xx 10^(-22) g` (iii) `1.295 xx 10^(22) g` (iv) `2.99 xx 10^(-23) g` (i) `6.022 xx 10^(23)` atoms of carbon have mass = 12.0 g 1.0 atom of carbon has mass `= (12.0)/(6.022xx10^(23))=1.99xx10^(-23)g` (ii) `6.022 xx 10^(23)` atoms of silver have mass = 107.8 g 1.0 atom of silver has mass `= (107.8g)/(6.022xx10^(23))=1.79xx10^(-22)g` (iii) `6.022 xx 10^(23)` molecules of benzene `(C_(6)H_(6))` have mass = 78.0 g 1.0 molecule of benzene `(C_(6)H_(6))` has mass `= (78.0g)/(6.022xx10^(23)) = 1.295 xx 10^(-22)g` (iv) `6.022xx10^(23)` molecules of `H_(2)O` have mass = 18.0 g 1.0 molecule of `H_(2)O` has mass `= (18.0 g)/(6.022 xx 10^(23))=2.99 xx 10^(-23) g`. |
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| 661. |
Multiply : (i) `730 xx 240` (ii) `5.02 xx 10^(23) xx 1.0064` (iii)`0.06204 xx 296.4` |
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Answer» Correct Answer - (i) `1.752 xx 10^(5)` (ii) `5.05 xx 10^(23)` (iii) 18.4 (i) `730xx240=175200=1.752xx10^(5)` (ii) `502xx10^(23)xx1.0064=5.05xx10^(23)` (iii) `0.06204xx296.4=18.39=18.4` (after rounded off) |
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| 662. |
`500 mL` of a glucose solution contains `6.02 xx 10^(22)` molecules. The concentration of the solution is :A. `0.1 M`B. `1.0M`C. `0.2M`D. `2.0M` |
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Answer» Correct Answer - C Moles of solute `= (6.02xx10^(22))/(N_(A)) = 0.1 mol` concentration of solution `= ("moles")/("vol")` `=(1xx1000)/(500) = 2` |
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| 663. |
What is the molality of a solution which contains 36 g of glucose `(C_(4)H_(12)O_(5))` in 250 g of water ? |
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Answer» Mass of glucose = 36 g Molar mass of glucose `= 6 xx 12+12xx1+6xx16=180 "amu"=180 " g mol"^(-1)` Mass of water `= 250 g = (250)/(1000)=0.25 kg` Molarity of solution `(m)=("Mass of glucose/Molar mass")/("Mass of water in kg")` `= ((36g))/((180" g mol"^(-1))xx(0.25 kg))=0.8"mol kg"^(-1)=0.8`m. |
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| 664. |
`N_(2)H_(4)`, Hydrazine a rocket fuel can be produced according to the following rection : `CINH_(2) + 2NH_(3) rarrN_(2)H_(4) + NH_(4)CI` When `1000 g CINH_(2)` is reacted with excess of `NH_(3),473 g N_(2)H_(4)` is produced. What is the `%` yield of the reaction. |
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Answer» Correct Answer - `76.12%` `ClNH_(2)+2NH_(3)rarrN_(2)H_(4)+NH_(4)Cl` `(1000)/(51.5)`mole excess `=19.417` `19.417` mole `%` yield `= (14.781)/(19.417)xx100=76.125%` |
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| 665. |
Carbon disulphide `CS_(2)` can be made from be product `SO_(2)`. The overall reaction is `5C+2SO_(2)rarr CS_(2)+4CO` How much `CS_(2)` can be produced from `450 kg` of waste `SO_(2)` with excess of coke kif the `SO_(2)` conversion is `82%`. |
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Answer» Correct Answer - `219.09kg CS_(2)` `5C+2SO_(2)overset(82%)(rarr) CS_(2) + 4CO` excess `(450)/(64) = 7.03K` mole `0.82xx(7.03)/(2) = 2.88 K` mole = 219.09kg` |
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| 666. |
Carbon disulphide, `CS_(2)` can be made from by-product `SO_(2).` The overall reaction is `5C+2SO_(2)rarrCS_(2)+4CO` How much `CS_(2)` (in kg) can be produced from 440 kg of waste `SO_(2)` with 60 kg of coke if the `SO_(2)` conversion is 80%? [Give answer after rounding off to the next integer.] |
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Answer» Correct Answer - 61 |
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| 667. |
A fluorine disposal plant was constructed to carryout the reactions: `F_(2)+2NaOHrarr1/2O_(2)+2NaF+H_(2)O` `2NaF+CaO+H_(2)OrarrCaF_(2)+2NaOH` As the plant operated, excess lime was added to bring about complete precipitation of the fluoride as `CaF_(2).` Over a period of operation, 1900 kg of fluorine was fed into a plant and 10,000 kg of lime was required. What was the percentage utilisation of lime? `[At, mass F=19], [Lime=CaO]` |
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Answer» Correct Answer - 28 |
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| 668. |
Calculate the molarity of `KCl` solution prepared by dissolving `7.45 g` of `KCl` in `500 mL` of the solution. `(d_(sol) = 1.2 g mL^(-1))` |
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Answer» Molar mass of `KCl = (39 + 35.5) g = 74.5 g` ` W_(2) 7.45 g, Mw_(2) = 74.5 g` `V_(sol) = 500 mL` `d_(sol) = 1.2 g mL^(-1)` `m = (W_(2) xx 1000)/(Mw_(2) xx W_(1))` In the above relation, `W_(1)` is known, so find `W_(1)`. `W_(1) = (W_(sol) - W_(2)) g = (V_(sol) xx d_(sol) - W_(2)) g` `(500 xx 1.2 - 74.5) g = 525.5 g` `:. m = (7.45 xx 1000)/(74.5 xx 525.5) = 0.19 mol kg^(-1) = 0.19 m` |
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| 669. |
Calculate the average atomic mass of hydrogen using the following data Isotope % Natural abundance Molar mass `.^(1)H 99.985 1` `.^(2)H 0.015 2` |
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Answer» Average atomic mass of hydrogen `= (("Natural abundance of" .^(1)"H" xx"atomic mass")+("Natural abundance of".^(2)"H"xx"atomis mass of".^(2)H))/(100)` `= (99.985xx1+0.015xx2)/(100)` `=(99.985+0.030)/(100)=(100.015)/(100)=1.00015u`. |
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| 670. |
Arrange the following in order of increasing masses. i. 1 molecules of oxygen ii. 1 atom of nitrogen iii. 1 mol of water iv. `1 xx 10^(-10)` of iron a. ii lt I lt iii lt iv b. ilt iilt ivlt iii c. ii lt ilt ivlt iii d. iltiiltiiiltiv |
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Answer» i. molecule of `O_(2) = (32)/(6 xx 10^(23))` i. a atom of `N = (14)/(6 xx 10^(23))` iii. 1 mol of `H_(2) O = 18 g` iv. Weight of `Fe = 10^(-18) g` So, the order is (c ). |
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| 671. |
Carbon occur in nature as a mixture of `C-12` and `C-13`. Average atomic mass of carbon is `12.011` what is the `%` abundance of `C-12` in nature ?A. `98.9%`B. `60.9%`C. `32.9%`D. `1.4%` |
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Answer» Correct Answer - 1 `{:("Isotopes of carbon",,C^(12),,C^(13)),(% " of abundance",=,x%,,(100-x)%):}` Average atomic mass `=(x xx12)/(100)+((100-x)xx13)/(100)` `12.011=(12x)/(100)+(1300-13x)/(100)` `12.011=(12x+1300-13x)/(100)=(-x+1300)/(100)` `12.011=(-x+1300)/(100)` `-x+1300=1201.1` `-x=12.011-1300` `-x=-98.9` `x=98.9%` `:. C^(12)=98.9%` `C^(13)=1.1%` |
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| 672. |
Rearrange the following `(I ` to `IV)` in the order of increasing masses `:` (I) `0.5` mole of `O_(3)" "(II) 0.5 gm` atom of oxygen (III) `3.011 xx 10^(23)` molecules of `O_(2) " " (IV) 5.6` litre of `CO_(2)` at `STP`A. `II lt IV ltIII lt I`B. `II lt Ilt IV ltIII`C. `IV lt II lt III lt I`D. `I lt II lt III lt IV` |
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Answer» Correct Answer - A `(I)" "0.5 mol e O_(3)=24gO_(3), " "(II)" "0.5g` at om of oxygen`=8g` `(III)" "(3.011xx10^(23))/(6.022xx10^(23))xx32=16gO_(2)," "(IV)" "(5.6)/(22.4)xx44gCO_(2)=11gCO_(3)` |
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| 673. |
Rearrange the following (P to S) in the order of increassing masses: (P) 0.5 mole of `O_(3)` (Q) 0.5gm molecules of nitrogen (R )`3.011 xx 10^(23)` molecule of `O_(2)` (S) 11.35 L of `CO_(2)` at STPA. `S ltR lt Q lt P`B. `Q lt R lt S lt P`C. `R lt Q lt P lt S`D. `P lt Q lt R lt S` |
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Answer» Correct Answer - B |
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| 674. |
In Haber process 30 litre of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only`50%` of the expected product. What will be the composition of gaseous mixture under the aforesaid condition in the end ?A. 20 litres `NH_3`, 25 litres `N_2`, 20 litres `H_2`B. 10 litres `NH_3`, 25 litres `N_2`, 15 litres `H_2`C. 20 litres `NH_3`, 10 litres `N_2`, 30 litres `H_2`D. 20 litres `NH_3`, 25 litres `N_2`, 15 litres `H_2` |
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Answer» Correct Answer - B |
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| 675. |
Balance the following chemical equation by partial equation method. `Cu+HNO_(3)rarrCu(NO_(3))_(2)+NO+H_(2)O` |
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Answer» (i) The probable partial equations for the above reaction are : `{:(HNO_(3)rarrH_(2)O+NO+[O]....("equation 1")),(Cu+OrarrCuO....("equation 2")),(Cu+HNO_(3)rarrCu(NO_(3))_(2)+H_(2)O....("equation 3")):}` (ii) After balancing by hit and trial method, the balanced partial equations may be written as : `{:(2HNO_(3)rarrH_(2)O+2NO+3O....("equation 1")),(Cu+OrarrCuO....("equation 2")),(CuO+2HNO_(3)rarrCu(NO_(3))_(2)+H_(2)O....("equation 3")):}` (iii) Multiply equation 2 as well as equation 3 by 3 in order to cancel out O and CuO which donot appear in the final equation. Finally add the three partial equations to get the final equation `{:(" "2HNO_(3)rarrH_(2)O+2NO+3O....("equation 1")),(" "Cu+OrarrCuO"]"xx3....("equation 2")),(" "Cu+2HNO_(3)rarrCu(NO_(3))_(2)+H_(2)O"]"xx3.....("equation 3")),(bar(3Cu+8HNO_(3)rarrCu(NO_(3))_(2)+2NO+4H_(2)O)):}`. |
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| 676. |
Equal mass of `KClO_(3)` undergoes different reactions in two different containers: `2KClO_(3)(s) overset(Delta)to 2KCl(s) + 3O_(2)(g)` ………………..(i) `4KClO_(3)(s) overset(Delta)to KCl(s) + 3KClO_(4)(s)`………………..(ii) Mass ratio of KCl produced in respective reaction is `n:1` then, value of n will be:A. 4B. 2C. 74.5D. 90 |
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Answer» Correct Answer - A |
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| 677. |
Excess of calcium orthophosphate is reacted with magnesium to form calcium phosphide `(Ca_(3)P_(2))` along with magnesium oxide. Calcium phosphide on reacting with excess of water liberate phosphine gas `(PH_(3))` along with calcium hydroxide. Phosphine is burnt in excess of oxygen to form `P_(2)O_(5)` along with water. Oxides of magnesium and phosphorous react to give magnesium metaphophate. Calculate grams of magnesium metaphosphate obtained if 1.92 gm of magnesium is taken. [Round off your answer to nearest integer] |
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Answer» Correct Answer - 2 |
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| 678. |
What is the maximum mass of `H_(2)O` (in gm) which can be obtained if total 42 gm of propyne and oxygen are subjected to combustion? |
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Answer» Correct Answer - 9 |
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| 679. |
Upon mixming `100.0 mL` to `0.1 M` potassium solphate solution and `100.0 mL` of `0.05 M` barium chloride solution, precipitation of barium sulphate takes place. How many moles of barium sulphate are formed? Also, calculate the molar concentration of species left behind in the solution. Which is the limiting reagent? |
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Answer» `K_(2) SO_(4) + BaCl_(2) rarr BaSO_(4) darr + 2 KCl` `(Acc.to equation), 1 "mmol, 1mmol, 1mmol, 2mmol")`, equation), mmole before, `10 xx 01, 100 xx 0.05`, -, - (reaction, = 10 mmol, = mmol), mmole left after, 10 - 5, -, 5 mmol, 10 mmol), reaction), = 5 mmol 5 mmol of `BaCl_(2)` will react will 5 mmol of `K_(2) SO_(4)`. Hence`BaCl_(2)` is the limiting reagent which is completely consumed in the reaction and it will decied the number of millimiles of products formed. Total volume of solution `= 100 + 100 = 200 mL` Moles of `BaSO_(4)` precipitated `= (5 mmol)/(1000 mL) = 0.005 mol` Unreacted `K_(2) SO_(4)` and `KCl` are left in the solution. Concentration of `K_(2) SO_(4)` and `KCl` are left in the solution `= (5 mmol)/(200 mL)` `= 0.025 M` Concentration of `KCl` formed in the reaction `= (10 mmol)/(200 mL)` `= 0.05 M` |
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| 680. |
`A" " 0.2g` sample , which is mixture of `NaCl, NaBr` and `NaI` was dissolved in water and excess of `AgNO_(3)` was added. The precipitate containing `AgCl,AgBr` and `AgI` was filtered, dried and weighed to be `0.412g`. The solid was placed in water and treated with excess of `NaBr`, which converted all `AgCl` into `AgBr`. The precipitate was then weighed to be `0.4881 g`. It was then placed into water and treated with excess of `NaI`, which converted all `AgBr` into `AgI`. The precipitate was then weighed to be `0.5868 g`. What was the percentage of `NaCl, NaBr` and `NaI` in the original mixture : |
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Answer» Correct Answer - `50,20.23,29.77` Working in backward direction In the last step moles of `(AgBr+Agl)`= moles of `AgI` `implies (0.4881-x)/(188)+(x)/(235)=(0.5868)/(235)impliesx=0.0933g` Mass `%` of `NaI` `=(0.0933)/(235)xx150xx(100)/(0.2)=29.77` Now subtracting mass of `AgI` from `1st` and `2nd` precipitate gives Mass of `(AgCl+AgBr)=0.3187 g` and mass of `AgBr=0.3948 g` Again `(y)/(143.5)+(0.3187-y)/(188)=(0.3948)/(188)impliesy=0.245g` implies Mass `%` of `NaCl` `=(0.245)/(143.5)xx58.5xx(100)/(0.2) = 50` Mass `%` of `NaBr = 20.23` |
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| 681. |
Which of the following conentration factor is affected by change in temperature?A. MolarityB. MolalityC. Mole fractionD. Weight fraction |
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Answer» Correct Answer - A |
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| 682. |
The mass of a single molecule of an allotrope of sulphur is `3.20 xx 10^(-22)`g. How many sulphur atoms are present in a molecular of this allotrope?A. 4B. 6C. 8D. 12 |
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Answer» Correct Answer - B |
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| 683. |
`0.45 g` of an orgainc compound containing only `C, H` and `N` on combustion gave `1.1 g` of `CO_(2)` and `0.3 g` of `H_(2) O`. What is the percentage of `C, H` and `N` in the orgainc compound. |
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Answer» Correct Answer - B::C::D `% of C = (12)/(44) xx ("Weight of" CO_(2) xx 100)/("Weight of compound")` `(12 xx 1.1 xx 100)/(44 xx 0.45) = 66.66%` `% of H = (2)/(18) xx ("Weight of" H_(2) O xx 100)/("Weight of compound")` `= (2 xx 0.3 xx 100)/(18 xx 0.45) = 7.4%` `% of N = [100 - (66.66 + 7.4)] = 25.93%` |
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| 684. |
The simplest formula of a compound containing 50% of an element `X` (atomic weight 10) and 50% of element `Y` (atomic weight 20) is:A. `XY`B. `X_(2)Y`C. `XY_(2)`D. `X_(2)Y_(3)` |
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Answer» Correct Answer - B `X : 50%` `Y : 50%` `X : Y = 5: 2.5 = 2 : 1,` hence `X_(2) Y` |
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| 685. |
Carbohydrates are compounds containing only carbon, hydrogen and oxygen having the atomic ratio of `H : O` as `2 : 1`. When heated in the absence of air, these compounds decompose to form carbon and water. a. If `310 g` of a carbohydrates leaves a residue of `124 g` of carbon on heating in absence of air, whatis the empirical formula of the carbohydrate? If 0.0833 mole of hte carbohydrate contains `1.0 g` hydrogen , what is the molecular formula of the carbohydrate? |
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Answer» Correct Answer - A::B::C a. Empirical formula : Carbon present `= 124 g` water present `= 310 - 124 = 186 g` Moles of Carbon `= (124)/(12) = 10.03` Moles of `H_(2) O = (186)/(18) = 10.03` Ratio of `C : H_(2) O = 1 :1` b. Molecular formula : 0.0833 moles of the carbohydrate contains `1.00 g` or one atom of hydrogen. Therefore, one mole of the carbohydrate will contains hydrogne atoms `= 1//0.0833 = 12` Number of hydrogen atoms in the molecules being 12, the molecular formula should be six times the empirical formula, i.e., `CH_(2) O xx 6` or `C_(6) H_(12) O_(6)` Hence, the molecular formula of the carbohydrate is `C_(6) H_(12) O_(6)`. |
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| 686. |
Carbohydrates are compounds containing only carbon, hydrogen and oxygen having the atomic ratio of `H : O` as `2 : 1`. When heated in the absence of air, these compounds decompose to form carbon and water. a. If `310 g` of a carbohydrates leaves a residue of `124 g` of carbon on heating in absence of air, what is the empirical formula of the carbohydrate? b. If 0.0833 mole of the carbohydrate contains `1.0 g` hydrogen , what is the molecular formula of the carbohydrate? |
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Answer» Step I. Calculation of the empirical formula of carbohydrate Percentage of carbon `= ("Mass of carbon")/("Mass of carbohydrate")xx100=(("124 g"))/(("310 g"))xx100=40` Percentage of water `= 100-40=60` `{:("Element/Compound","Percentage","Atomic mass (Molar mass)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("Carbon (C)",40,12,(40)/(12)=3.3,1),("Water"(H_(2)O),60,18,(60)/(18)=3.33,1):}` Empirical formula : `C_(1)(H_(2)O)_(1) or CH_(2)O` Step II. Calculation of the molecular formula of carbohydrate 0.0833 mole of carbohydrate contains H = 1.0 g or 1 atom 1 mole of carbohydrate contains H atom `= (("1 atom"))/(("0.0833 mol"))xx"1 mol"=12` atoms Thus, 1 mole and also one molecule of carbohydrade contain hydrogen atoms = 12 Since hydrogen and oxygen are present in the ratio of 2 : 1 therefore, 1 molecule of carbohydrate contain oxygen atoms = 6 `:.` Molecular formula of carbohydrate `= (CH_(2)O)xx6 = C_(2)H_(12)O_(6)`. |
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| 687. |
The general formula of a carbohydrate is `C_(x)(H_(2)O)_(y)`. 3.1 g of the carbohydrate on heating in the absence of oxygen yields 1.24 g of carbon. The molecular mass of carbohydrate is 180u. Calculate its molecular formula. |
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Answer» The carbohydrate may be regarded as the hydrate of carbon Percentage of carbon (C) in carbohydrate `= ((1.24g))/((3.10g))xx100=40` Percentage of water `(H_(2)O)` in carbohydrate `= 100 - 40 = 60` Moles of `C=((40 g))/(("12 g mol"^(-1)))=3.33` mol Moles of `H_(2)O=((60g))/(("18 g mol"^(-1)))=3.33` mol `:.` Mole ratio of `C :H_(2)O : : 1 : 1` Empirical formula of carbohydrate `= C (H_(2)O)` Empirical formula mass `= 12 + 18 = 30` u Molecular mass = 180u (given) , `n = ("Molecular mass")/("Empirical formula mass")=(("180 u"))/(("30 u"))=6` `:.` Molecular formula of carbohydrate `= 6 xx (CH_(2)O)=C_(6)H_(12)O_(6)`. |
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| 688. |
One of the reactions used in the petroleum industry for improving octance number of fuels is `C_(7) H_(14) rarr C_(7)H_(8) + 3H_(2)` The two hydrocarbons `C_(7) H_(14)` and `C_(7)H_(8)` are liquid, `H_(2)` formed is gas. Whatis the percentage reduction in liquid weight accompanying the completion of the above reaction?A. `~~ 1%`B. `~~ 3%`C. `~~ 5%`D. `~~ 6%` |
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Answer» Correct Answer - D Ratio `= (1 "mol" C_(7) H_(8) (92.0 g//"mol"))/(1 "mol" C_(7)H_(17) (98.0 g//"mol")) = 0.939` % reduction `= ((1 - 0.939)/(1)) xx 100 = 6.1%` |
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| 689. |
A metal oxide has the formul `Z_(2)O_(3)`. It can be reduced by hydrogen to give free metal and water. `0.2 g` of the metal oxide requires `12 mg` of hydrogen for complete reduction. The atomic weight of the metal isA. 52B. 104C. 26D. 78 |
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Answer» Correct Answer - C `underset([2g = 200 mg])(Z_(2) O_(3)) + underset({:((12 mg)/(2 mg)),(= 6 "mol"):})(3H_(2)) rarr 2Z + 3H_(2) O` Since `H_(2)` used is `6 m` ml, `Z_(2) O_(3)` used should be ` 2 "mmol"`. Mol of `Z_(2) O_(3) = ("Weight")/(Mw) = (200 mg xx 10%(-3) g)/(Mw)` `= 2 x 10^(-3)` moles `:. Mw (Z_(2) O_(3)) = 100` `:. 2Z + 16 xx 3 = 100` `Z = (100 - 48)/(2) = 26` |
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| 690. |
Which has the maximum number of atoms ?A. `6.022 xx 10^(21)` molecules of `CO_(2)`B. 22.4 L of `CO_(2)` at N.T.PC. 0.44 of `CO_(2)`D. None of these |
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Answer» Correct Answer - B 22.4 L of `CO_(2)` (1 gram mole) has maximum no. of carbon atoms. |
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| 691. |
`10.0 L` of air of `STP` was slowly bubbled through `50 mL` of `N//25Ba(OH)_(2)` solution and the final solution rendered red with phenoophthalein. After filtering the solution from the precipitated `BaCO_(3)`, the filtrate requried `22.5 mL` of `N//12.5 HCl` to becomes just colourless. Calculate the % age by volume of `CO_(2)` in the air. |
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Answer» Air contains `CO_(2)` (acidic oxide) which reacts with `Ba(OH)_(2)` as: `Ba(OH)_(2) + CO_(2) rarr BaCO_(3) + H_(2) O` It is clear from the data the `CO_(2)` in the air is not sufficient to neutrialise whole of hydroxide, so `HCl` is used to completely neutralise the solution. mEq of `Ba(OH)_(2) = (1)/(12.5) xx 22.5 = 1.8` Initial mEq of `Ba (OH)_(2) = (1)/(25) xx 50 = 2` mEq of `Ba(OH)_(2)` neutralised by `CO_(2)` in the air `= 2 - 1.8 = 0.2` As `Ba(OH)_(2)` is diacidic base: 1 mEq of `Ba(OH)_(2) = 1//2` millimoles of `Ba(OH)_(2)` `implies` Millimoles of `Ba(OH)_(2) = 0.2//2 = 0.1` `implies` Moles of `Ba(OH)_(2) = 0.1 xx 10^(-3) = 10^(-4)` moles From stoichiometry, we have: 1 mol of `Ba(OH)_(2) -=` 1 mole of `CO_(2)` `implies` Moles of `CO_(2) = 10^(4)` `implies` volume at `STP = 10^(-4) xx 22.4 L` Percentage of `CO_(2) = (22.4 xx 10^(-4))/(10) xx 100 = 0.0224 %` |
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| 692. |
The mass of `CaCO_(3)` produced when carbon dioxide is bubbled through 500 mL of 0.5 M `Ca(OH)_(2)` solution will be :A. 25 gB. 50 gC. 20 gD. 10 g |
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Answer» Correct Answer - A Mass of `Ca(OH)_(2)` is calculated as follows : Molarity `= (("Mass of "Ca(OH)_(2))/("Molecular mass"))/("Volume of solution in litres")` `0.5=("Mass of "Ca(OH)_(2))/(74)xx(1000)/(500)` `=0.5xx39 = 18.5g` `{:(Ca(OH)_(2)+CO_(2),rarr,CaCO_(3)+H_(2)O),(74g,,100g),(18.5g,,(100)/(74)xx18.5=25g):}`. |
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| 693. |
One gram of a metallic oxide on reduction gives 0.68 g of the metal. The equivalent mass of the metal is :A. 68B. 34C. 51D. 17 |
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Answer» Correct Answer - D Equivalent mass of metal `= ("Mass of metal")/("Mass of oxygen")xx8=(0.68)/(0.32)xx8=17`. |
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| 694. |
From `200 mg` of `CO_(2), 10^(21)` molecules are removed. How many grams and moles of `CO_(2)` are left. |
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Answer» `44 g of CO_(2) = 1 mol = 6.023 xx 10^(23)` molecules `:. 6.023 xx 10^(23)` molecules `= 44 g of CO_(2)` `10^(21)` molecules `= (40 xx 10^(21) xx 10^(3))/(6.023 xx 10^(23)) mg` Weight of `CO_(2)` left `= 200 - 73.05 = 126.9 mg` `(126.9)/(10^(3)) = 0.1269 g` `44 g of CO_(2) = 1 mol` `0.1269 g of CO_(2) = (1)/(44) xx 0.1269 = 0.0028 mol` |
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| 695. |
If 500 mL of a 5M solution is diluted to 1500 mL, what will be molarity of the solution obtained ?A. 1.5 MB. 1.66 MC. 0.017 MD. 1.59 M |
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Answer» Correct Answer - B `M_(1)V_(1)-=M_(2)V_(2)` `(5M)xx(500mL)-=M_(2)xx(1500mL)` `M_(2)=((5M)xx(500mL))/((1500mL))=1.66M` |
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| 696. |
Comprehension # 8 `A` factory, producing methanol, is based on the reaction: `CO+2H_(2)hArrCH_(3)OH` Hydrogen `&` carbon monoxide are obtained by the reaction `CH_(4)+H_(2)OhArrCO+3H_(2)` Three units of factory namely, the "reformer" for the `H_(2)` and `CO` production, the "methanol reactor" for production of methanol and a "separator" to separate `CH_(3)OH` from `CO` and `H_(2)` are schematically shown in figure. four positions are indicated as `alpha,beta,gamma` and `delta`. The flow of methanol at position `gamma` is `10^(3)mol//sec`. The factory is so designed that `(2)/(3)` of the `CO` is converted to `CH_(3)OH`. Excess of `CO` and `H_(2)` at position `delta` are used to heat the first reaction. Assume that the reformer reaction goes to completion. At the position `(beta)` mole ratio of `CO` to `H_(2)` is `(1)/(3)` `CO+2H_(2)hArrCH_(3)OH " "DeltaH_(r)=-100 R` Amount of energy released in methanol reactor in `1` minute ?A. `1200 kcal`B. `12000 kcal`C. `6000 kcal`D. None of these |
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Answer» Correct Answer - B `1000xx60xx2xx100 = 12000 Kcal` |
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| 697. |
Comprehension # 8 `A` factory, producing methanol, is based on the reaction: `CO+2H_(2)hArrCH_(3)OH` Hydrogen `&` carbon monoxide are obtained by the reaction `CH_(4)+H_(2)OhArrCO+3H_(2)` Three units of factory namely, the "reformer" for the `H_(2)` and `CO` production, the "methanol reactor" for production of methanol and a "separator" to separate `CH_(3)OH` from `CO` and `H_(2)` are schematically shown in figure. four positions are indicated as `alpha,beta,gamma` and `delta`. The flow of methanol at position `gamma` is `10^(3)mol//sec`. The factory is so designed that `(2)/(3)` of the `CO` is converted to `CH_(3)OH`. Excess of `CO` and `H_(2)` at position `delta` are used to heat the first reaction. Assume that the reformer reaction goes to completion. At the position `(beta)` mole ratio of `CO` to `H_(2)` is `(1)/(3)` `CO+2H_(2)hArrCH_(3)OH " "DeltaH_(r)=-100 R` What is the flow of `CO` and `H_(2)` at position `(beta)` ?A. `CO : 1500mol//sec., H_(2) : 2000 mol//sec`.B. `CO : 1500mol//sec., H_(2) : 3000 mol//sec`.C. `CO : 1000mol//sec., H_(2) : 2000 mol//sec`.D. `CO : 1500mol//sec., H_(2) : 4500 mol//sec`. |
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Answer» Correct Answer - D `{:(,CH_(4)+H_(2)OrarrCO,+,3H_(2)),(," "(3)/(2)xx10^(3),,3):}` `CO+2H_(2) rarr CH_(3)OH` `10^(3)" "2xx10^(3)" "10^(3)("moles/sec")` `("total" CO = (3)/(2) xx 10^(3))` |
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| 698. |
Comprehension # 8 `A` factory, producing methanol, is based on the reaction: `CO+2H_(2)hArrCH_(3)OH` Hydrogen `&` carbon monoxide are obtained by the reaction `CH_(4)+H_(2)OhArrCO+3H_(2)` Three units of factory namely, the "reformer" for the `H_(2)` and `CO` production, the "methanol reactor" for production of methanol and a "separator" to separate `CH_(3)OH` from `CO` and `H_(2)` are schematically shown in figure. four positions are indicated as `alpha,beta,gamma` and `delta`. The flow of methanol at position `gamma` is `10^(3)mol//sec`. The factory is so designed that `(2)/(3)` of the `CO` is converted to `CH_(3)OH`. Excess of `CO` and `H_(2)` at position `delta` are used to heat the first reaction. Assume that the reformer reaction goes to completion. At the position `(beta)` mole ratio of `CO` to `H_(2)` is `(1)/(3)` `CO+2H_(2)hArrCH_(3)OH " "DeltaH_(r)=-100 R` What is the flow of `CO` and `H_(2)` at postion `(gamma)` ?A. `CO:500 mol//sec.,H_(2):1000mol//sec`.B. `CO:500 mol//sec.,H_(2):2500mol//sec`.C. `CO:500 mol//sec.,H_(2):2000mol//sec`.D. `CO:500 mol//sec.,H_(2):1500mol//sec`. |
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Answer» Correct Answer - B `CO = 500 m//s " "H_(2) = 2500 "mole/sec"` left |
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| 699. |
`20 mg` of `K^(o+)` ions are present in `1 L` of aqueous solution. Density of the solution is `0.8 mL^(-1)`. What is the concentration of `K^(o+)` ions in ppm? |
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Answer» Mass of `K^(o+)` ions `(W_(2)) = 20 mg = 20 xx 10^(-3) g` Mass of solution `= V_(sol) xx d_(sol) = 100 mL xx 0.8 g mL^(-1) = 800 g` `ppm = (W_(2) xx 10^(6))/(W_(sol)) = (20 xx 10^(-3) xx 10^(6))/(800) = 25 ppm` |
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| 700. |
The total charge in coulombs required to complete the electrolysis is :A. 24125B. 48250C. `96500`D. `193000` |
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Answer» Correct Answer - D `Na^(+) + e^(-) rarr Na` Moles of Na discharged at cathode = 1 mol Total charge needed to complete the electrolysis is `= 2 xx F = xx 96500 C` `= 193000 C`. |
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