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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.
| 7051. |
Can the gas produced by the reaction between ethanoic acid and sodium carbonate be identified by this test ? |
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Answer» SOLUTION :YES. Carbon dioxide gas PRODUCED by the reaction can be identified by this TEST. Repeat this activity with sodium hydrogen CARBONATE instead of sodium carbonate. |
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| 7052. |
Can the following group of elements be classified as Dobereiner's Triad ? Na,Si,CI |
| Answer» SOLUTION :No, These ELEMENTS do not have SIMILAR CHEMICAL PROPERTIES. | |
| 7053. |
Can sodium metal be cut with a knife ? |
| Answer» SOLUTION :YES, we can cut sodium metal into PIECES. | |
| 7055. |
Can plaster of Paris be obtained by heating calcium sulphate at a temperature beyond control ? |
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| 7056. |
Can onion juice be used by a blind student to test whether a solution is acidic or basic ? |
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| 7057. |
Can Na_(2)CO_(3) be used in place of NaOH in the preparation of the soap ? |
| Answer» Solution :No , because `Na_(2)CO_(3)` does not contain -OH group to participate in the REACTION . | |
| 7058. |
Can dry hydrogen chloride gas turn a blue litmus paper red ? |
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| 7059. |
Can copper displace zinc or lead from their salt solutions? |
| Answer» SOLUTION :No,COPPER DISPLACE zinc or lead their SALT SOLUTIONS .Because copper is less reactive than zinc and lead. | |
| 7061. |
Can a nonmetal displace hydrogen from an acid ? |
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| 7062. |
Can a nickel spatula be used to stir sulphate solution?Justify your answer. |
| Answer» SOLUTION :No, nickel spatula cannot be USED to STIR COPPER sulphate solution.Actually on the basis of activity series ,nickel is more reactive than copper,so nickel displace copper from its solution and copper will be deposited on nickel spatula. | |
| 7063. |
Can a double displacement reaction take place when the products are highly soluble or highly ionised? |
| Answer» Solution :No, DOUBLE displacement reaction takes PLACE when there is a FORMATION of a slightly SOLUBLE salt. | |
| 7064. |
Can a displacement reaction be a redos reaction ? Explain with the help of an example . |
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Answer» Solution :Yes, all displacement reactions are redox reactions. Consider the reaction of SODIUM with water , in this reaction sodium displaces HYDROGEN from water, it is ALSO GETTING oxidized by gaining oxygen, and simultaneously reducing the water to hydrogen by removal of oxygen. `2Na+H_(2)O to Na_(2)O+H_(2)`. |
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| 7065. |
Can a detergent be used to produce lather when the water is hard ? |
| Answer» Solution :Yes, a detergent can PRODUCE lather EVEN when the WATER is HARD. | |
| 7066. |
Calculated the difference in the formulae andmolecular masses for (a) CH_3OH and C_2H_5OH (b) C_2H_5OH and C_3H_7OH and (c) C_3H_7OH and C_4H_9OH • Is there any similarity in these three ? • Arrange these alcohols in the order of increasing carbon atoms to get a family . Can we call this family a homologous series ? |
| Answer» Solution :The difference in the molecular formulae is `-CH_2` i.e., 14 amu. All the three show the same difference. The INCREASING order in terms of a carbon atoms is `CH_3OH, C_2H_5OH, C_3H_7OH` and `C_4H_9OH`. YES, we we can call this as homologous SERIES. | |
| 7067. |
Calculate the weight of zinc required for the liberation of 10 g of hydrogen gas on reaction with H_(2)SO_(4). |
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Answer» Solution :`{:(Zn+H_(2)SO_(4) toZnSO_(4) +H_(2)),("1 MOLE""1 mole"):}` Atomic WEIGHT of zinc is ` 65.5` 1 mole of HYDROGEN = 2 g of hydrogen 10 g of hydrogen = `10/2 = 5` moles 1 mole of `H_(2)` gas is produced FROM1 mole of zinc 5 moles of `H_(2)` gas are produced from 5 molesof zinc. 1 mole of zinc = 65.5g 5 mole of zinc = ` 65.5 xx 5 = 327. 5 ` g |
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| 7068. |
Calculate the weight of sodium bicarbonate to be dissociated to give 0.56 L of CO_(2) gas. |
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Answer» SOLUTION :`{:(2NaHCO_(3) toNaCO+HO+CO),("2 moles""1 mole"):}` GMW of `NaHCO_(3)` = 84 g 1 mole of `CO_(2)` occupies22.4 L `22 4" of " CO_(2)` is produced from` 2 xx 84" g of "NaHCO_(3)` ` 0.56 " L of "CO_(2)` is produced from `(0.56 xx 2 84)/(22.4) xx 4.2` g |
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| 7069. |
Calculate the weight of 80% pure limestone required to produce 11 g of CO_(2) gas. |
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Answer» Solution :`{:(CaCO_(3) to CaO+CO_(2)),(" 1 MOLE""1 mole"):}` 44 g of `CO_(2)` is produced from 100 g of `CaCO_(3)` `:. 11" g of "CO_(2)` is produced from` 100/44 xx 11 = 25" g of "CaCO_(3)` As thelimestone is 80% pure, THEWEIGHT of IMPURE LIMESTONE required = `100/80 xx 25" g " = 31.2` g. |
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| 7070. |
Calculatethe Volumeof ethanolin 200 mlsolutionof 20 %v/vaqueoussolutionof ethanol . |
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Answer» SOLUTION :Volumeof aqueoussolution= 200 ml Volumepercentage= 20 % Volumepercentage`= (Volume of thesolute ")/("Volumeof THESOLUTION ")xx 100 ` `20=("Volumeof ethanol ")/("200 ") xx 100` Volumeof ethanol`=( 20 xx 200)/(100) = 40 ml` |
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| 7071. |
Calculate the volume of carbon monoxide gas required to react with oxygen to give 11.2 L of CO_(2) gas. |
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Answer» Solution :`{:(2CO+O_(2) to 2CO_(2)),(" 2 MOLES""2 moles"):}` `2 xx 22.4 "L"CO_(2)` gas is PRODUCED from `2 xx 22.4` L CO gas. `11.2" L " CO_(2)` gas is producedfrom11.2 L CO gas. |
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| 7072. |
Calculate the volume occupied by : 3.011xx10^(23) of ammonia gas molecules |
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Answer» Solution :`"Number of MOLES "=("Nuber of molecules")/("Avogadro.s number")` `=3.011xx10^(23)//6.023xx10^(23)` `="2 moles"` `"VOLUME occupied by "NH_(3)="number of moles"XX"molar volume"` `=2xx22.4` `"44.8 LITRES at S.T.P"` |
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| 7073. |
Calculate thevolume occupied by200 g of SO_(3) gas at STP and the number ofmolecules present in it. |
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Answer» Solution :Weight of `SO_(3)` taken = 200 G Number of moles of `SO_(3) = 200/("GMW of "SO_(3))=200/80 = 2.5 ` moles Volume occupied by 1 mole of gas is22.4 L, at STP Volume occupied by thegiven AMOUNT of gas = `2.5 xx 22.4 = 56` L Number of molecules PRESENT in 1mole of gas= ` 6.023 xx 10^(23)` Number of molecules present in thegiven amount of gas= `2.5 xx 6.023 xx 10^(23) = 15.05 xx 10^(23)` molecules |
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| 7074. |
Calculate the volume occupied by : "1.5 mole of "CO_(2)" at S.T.P" |
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Answer» Solution :`"Number of MOLES of "CO_(2)=("GIVEN VOLUME of S.T.P")/("Molar volume at S.T.P")` `"2.5 mole of "CO_(2)=("Volume of "CO_(2)" at S.T.P")/(22.4)` `"Volume of "CO_(2)" at S.T.P "=22.4xx2.5="56 litres."` |
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| 7075. |
Calculate the volume occupied by : 14 g nitrogen gas |
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Answer» Solution :`"NUMBER of moles"=11//28=0.5" moles"` `"Volume OCCUPIED by "N_(2)" at S.T.P"="No. of moles"XX"MOLAR volume"` `=0.5xx22.4` `="11.2 litres"` |
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| 7076. |
Calculate the total number of electrons present in 1.6 g of methane. |
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Answer» Solution :(i) Molar MASS of METHANE `(CH_(4))=12+4xx1=16g` 16 g of methane contain molecules `=6.022xx10^(23)` `"1.6 g of methane contain molecule"=(6.022xx10^(23))/((16g))xx(1.6g)=6.022xx10^(21)" molecules."` (ii) Numberof electrons in `6.022xx10^(22)` molecules of methane 1 molecule of methane contains electrons `=6+4=10` `6.022xx10^(22)` molecules of methane contain electrons `=6.022xx10^(22)xx10=10=6.022xx10^(23)` |
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| 7077. |
Calculate the table given below. |
Answer» SOLUTION :
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| 7078. |
Calculate the standard enthalpy of formation of CH_(3)OH (I) from the following data: CH_(3)OH (I) + 3//2O_(2) (g) rarr CO_(2) (g) + 2H_(2)O(I) , Delta_(c ) H^(theta) = - 726kJ mol^(-1) C(g) + O_(2)(g) rarr CO_(2) (g), Delta_(f) H^(theta) = - 393 kJ mol^(-1) H_(2)(g) + 1//2O_(2) (g) rarr H_(2)O (I), Delta_(f) H^(ө) - 286kJ mol^(-1) |
| Answer» SOLUTION :`-239 KJ MOL^(-1)` | |
| 7079. |
Calculate the resultant pH of a solution when 20 mL 0.1 N NaOH is mixed with 20 mL of 0.05 M Ca(OH)_(2) " at " 25^(@)C |
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Answer» Solution :(i) CALCULATION of NET `H^(+)` ion concentration in the mixture (ii) calculation ofthe total NUMBER of `OH^(-)` ions generated from the given volume of NAOH and `Ca(OH)_(2)` (iii) calculation of concentrationof `OH^(-)` ions in the RESULTANT solution (iv) calculation of pOH (v) calculation of pH from the value of pOH at `25^(@)C` (vi) 3 |
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| 7080. |
Calculate the % relative abundance of B-10 and B-11, if its average atomic mass is 10.804 amu. |
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Answer» Solution :Average ATOMIC mass of Boron = 10.804 amu. `%" relative abundance of "B-10=?` `%" relative abundance of "B-11=?` Let the FRACTION of relative abundance of `B-10=x` Let the fraction of relative abundance of `B-11=y` `x+y=1` `y=1-x` `"Relative abundance"=x(10)+(1-x)(11)=10.804amu` `RARR 10x+11-11x="10.804 amu"` `11-x="10.804 amu"` `-x=10.804-11` `-x=-0.196` `x=0.196` `x=%" abundance of "B-10=0.196xx100=19.6%` `y=%" abundance of "B-11=100-19.6=80.4%` Percentage abundance of `B - 10 = 19.6%` Percentage abundance of `B-11=80.4%` |
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| 7081. |
Calculatethe %relativeabundanceofB-10and B-11, ifitsaverage atomicmassis 10 .804amu . |
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Answer» SOLUTION :Let%Abundanceof `B-10 = X ` %of relativeabundanceof `B - 11 =(100 - x)` Averageatomicmassof B `=(10xx x+11 (100-x))/(100)` `10.804 =(10x +1100- 11 x )/( 100 )` `10.804 xx 100 = 1100 - x ` `1080.4 = 1100-x ` ` x=1100 - 1080.4 ` ` thereforex= 19.6 % ` ` THEREFORE ` %abundanceof b- 10is 19.6 % ` therefore ` % Abundanceof B-11= (100 -x) `=100-19.6 ` `=80.4` |
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| 7082. |
Calculate the pH value of the 10^(-6) M HCl solution when diluted by 100 times. Justify (log 1.1 = 0.041) |
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Answer» Solution :`10^(-6)`M HCl solution when diluted by 100 times, should become `10^(-8)` M solution. Then the pH value should increase from 6 to 8. However, a solution having pH value of 6 is acidic and the solution having pH value of 8 indicates BASIC nature. An acidic solution can never become neutral or basic howsoever dilute it is. The pH value of the solution is slightly less than 7. This is because in very low concentrated solution, the dissociation of water also should be taken into consideration. `H^(+)` ion concentration should be equal to the sum of `H^(+)` ion concentration from HCl and `H_(2)O` `[H^(+)] " from " HCl = 10^(-8)` `[H^(+)] " from " H_(2)O = 10^(-7)` `[H^(+)] = 10^(-8) + 10^(-7) = 10^(-7) (10^(-1) + 1) = 1.1 XX 10^(-7)` `pH = -"LOG" (1.1 xx 10^(-7))` `= 7 - " log " 1.1 = 7 - 0.041 = 6.96` |
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| 7083. |
Calculate the pH of sodium hydroxide solution having the concentration of OH^(-) 0.01m L^(-1). |
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Answer» Solution :POH=`-log_(10)[OH^(-)]` pOH=`-log_(10)[1xx10^(-2)]` pOH=2 pH+pOH=14 `THEREFORE` pH=14-pH=14-2 pH=12 |
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| 7084. |
Calculate the pH of a solution in which the concentration of the hydrogen ions is 1.0xx10^(-8) mol litre^(-1) |
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Answer» Solution :Here ,although the solution is extremenly dilute ,the concentration given given is not an ACID or a base but that of `H^(+)` ions.Hence ,the PH can be calculated from the RELATION: pH=`-log_(10)10^(-8)=-(-8log_(10)10)` `=-(-8xx1)=8` |
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| 7085. |
Calculate the pH of 1xx10^(-4) molar solution of NaOH. |
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Answer» Solution :NAOH is a strong BASE and dissociates in its solution as: `NaOH_((aq))toNa_((aq))^(+)+OH^_(aq)^(-)` One mole of NaOH would one mole of `OH^(-)` ions.Therefore,`[OH^(-)]=1xx10^(-4)mol`,`litre^(-1)` pOH=`-log_(10)[OH^(-)]=-log_(10)xx[10^(-4)]` pH=14-pOH=14-4 10 |
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| 7086. |
Calculate the pH of 1.0xx10^(-4) molar solution of HNO_(3) |
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Answer» Solution :`HNO_(3)TOH^(+)NO_(3)^(-)` `1XX10^(-4)` molar`1xx10^(-4)` molar pH=-log[H+]=-log`[1xx10^(-4)]` PH=4 |
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| 7087. |
Calculatethe pHof 1.0 xx10 ^(-4) molar Solution ofHNO_(3). |
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Answer» Solution :`[H^+]= 1.0xx10^(-4)` `pH =- LOG _(10) [H^+]` `=- log _(10) [1xx10^(-4)]` `pH =-(log _(10)1-4log _(10)log _(10)^(10))` `=0 +4xx log _(10)^(10)` `=0+4xx1=4` `pH = 4 ` |
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| 7088. |
Calculate the pH of 0.01 M HNO_(3) ? |
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Answer» SOLUTION :`[H^(+)]`=0.01 pH=`log_(10)[H^(+)]` pH+`-log_(10)[1XX10^(-2)]` pH=`-(log_(10)1-2 log_(10)10)` pH=0+2`xx1=2` pH=2 |
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| 7089. |
Calculate the pH of 0.001 molar solution of HCI. |
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Answer» Solution :HCI is a strong acid and is completely dissociated in its solutions according to the process: `HCl_(aq)toH_(aq)^(+)+Cl_(aq)^(-)` From this process it is clear that one mole of HCI would give on mole of `H^(+)` ions. Therefor ,the concentration of `H^(+)` ions would be equla to that of HCl ,i.e., 0.001 molar or `1.0xx10^(10^(-3))` MOL `LITRE^(-1)` THUS `[H^(+)]=log_(10)10^(-3)` `=-(-3xxlof_(10))=-(3xx1)=3` Thus ,pH=3 |
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| 7090. |
Calculate the percentage of N in urea. (Molar mass of urea = 60" gmol"^(-1) ) |
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| 7091. |
Calculate the percentage composition of oxygen and hydrogen by taking the example of H_(2)O. |
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Answer» Solution :`"Mass % of an ELEMENT"=("Mass of that element in the compound")/("Molar mass of the compound")xx100` `"Now, Molar mass of "H_(2)O=2(1)+16=18g` `"Mass % of HYDROGEN"=(2)/(18)xx100=11.11%` `"Mass % of Oxygen"=(16)/(18)xx100=88.89%` |
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| 7092. |
Calculate the percent of carbon, hydrogen and oxygen in ethanol (C_(2)H_(5)OH) |
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| 7093. |
Calculate the partial pressures of each gas present in a mixture of 8 g of oxygen and6 g of hydrogen, present in 2 L container at 27^(@) C . |
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Answer» Solution :Number of moles of oxygen , `n_(O_(2)) = 8/32 = 0.25` Number of moles of hydrogen, `n_(H_(2)) = 6/3 = 2` Total number of moles, `n = 0.25 + 2 = 2.25 PV= nRT` ` 2P = 2.25 xx 0.08 xx 300` P = 27atm Partial pressure of oxygen, `pO_(2) ` = (mole FRACTION of`O_(2)`.P ` = ((0.25)/(0.25 + 2)) 27 = 3` atm `P_("total") = P_(O_(2)) + P_(H_(2))` Partial pressure of `H_(2) = 27 - 3 = 24` atm |
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| 7094. |
Calculate the % of oxygen in Al_(2)(SO_(4))_(3). ("Atomic mass: "Al-12, O-16, S-32) |
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Answer» Solution :`"Aluminum Sulphate "-Al_(2)(SO_(4))_(3)` `"Molar mass of Aluminium Sulphate"=(27xx2)+(32xx3)+(16xx12)` `=54+96+192=342g` `"% of OXYGEN "=(192)/(342)xx100=56.14%` |
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| 7095. |
Calculate the % of each element in calcium carbonate. (Atomic mass : C-12, O-16, Ca-40) |
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Answer» Solution :`"Calcium carbonate : "CaCO_(3)` `"Molar mass of "CaCO_(3)=40+12+(16xx3)=100G` `"% of Calcium"=(40)/(100)xx100=40%` `"% of CARBON "=(12)/(100)xx100=12%` `"% of Oxygen"=(48)/(100)xx100=48%` |
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| 7096. |
Calculate the % of each element in calcium carbonate. (Atomic mass: C-12, O-16, Ca-40) |
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Answer» Solution :Molarmassof `CaCO_3=40 +12+(3xx16)` = 100 %of ANELEMENT `=( "Massof anelementin thecompound ")/(" Molarmassof compound ")xx100` %ofCin `CaCO_3 =(12)/(100)xx100 =12% ` % ofO in `CaCO_(3) =(16xx3)/(100)xx100= 48 % ` %of Cain `CaCO_(3) =(40 )/(100)xx100 = 40 % ` |
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| 7097. |
Calculate the number of water molecule present in one drop of water which weighs 0.18 g. |
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Answer» SOLUTION :`"The molecular MASS of water "(H_(2)O)` is 18. 18 g of water molecule = 1 mole. `therefore"0.18 g of water"=(1)/(18)xx0.18="0.10 mole."` `"1 mole of water (Avogaro.s number) contains "6.023xx10^(23)" water molecules."` `therefore 0.01"mole of water contain "(6.023xx10^(23))/(1)xx0.01=6.023xx10^(21)" molecules."` |
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| 7098. |
Calculate the number of moles of a sample that contains 12.046xx10^(23) atoms of iron? |
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Answer» SOLUTION :`"NUMBER of moles "=("Number of atoms of IRON")/("Avogadro.s number")` `=12.046xx10^(23)//6.023xx10^(23)` `="2 moles of iron"` |
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| 7099. |
Calculate the number of moles in (i) 81 of Aluminium (ii) 4.6 g of sodium (iii) 5.1 g of ammonia (iv) 90 g of water (v) 2g of NaOH. |
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Answer» Solution :`"No of moles"=("Given mass")/("ATOMIC mass")` (i) `"No. of moles of Aluminium "=(81)/(27)` `="3 moles of aluminium"` (ii) `"No. of moles of Sodium"=(4.6)/(23)` `="0.2 moles of sodium"` (iii) No. of moles of AMMONIA `=(5.1)/(17)` = 0.3 moles of ammonia (IV) No. of moles of Water `=(90)/(18)` = 5 moles of water (v) No. of moles of `NaOH=(2)/(40)` = 0.05 moles of `NaOH` |
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| 7100. |
Calculate the number of moles in (i) 27 g Al (ii) 1.51xx10^(23)" molecules of "NH_(4)Cl |
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Answer» SOLUTION :(i) 27 g of Al `"Number of moles"=("Mass")/("ATOMIC mass")` `"Number of moles in 27 g of Al"=(27)/(27)="1 mole"` (ii) `1.51xx10^(23)" MOLECULES of "NH_(4)Cl` `"Number of moles"=("Number of molecules")/("Avogadro.s number")=(1.51xx10^(23))/(6.023xx10^(23))="0.25 moles."` |
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