InterviewSolution
Saved Bookmarks
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.
| 7101. |
Calculate the number of moles in (i) 12.046xx10^(23)" atoms of copper" (ii) 27.95 g of iron (iii) 1.51xx10^(23)" molecules of "CO_(2) |
|
Answer» SOLUTION :(i) `12.046xx10^(23)" atoms of copper"` `6.023xx10^(23)" atoms of copper = 1 mole"` `therefore 12.046xx10^(23)" atoms of copper"=(1xx12.046xx10^(23))/(6.023xx10^(23))="2 moles of copper"` (ii) `"27.95 g of iron"` `"55.9 g iron = 1 mole"` `therefore" 27.95 g of iron"=(1)/(55.9)xx27.95="0.5 mole of iron"` (iii) `1.51xx10^(23)" MOLECULE of "CO_(2)` `"No. of moles"=("No. of MOLECULES")/("Avogadro NUMBER")` `=(1.51xx10^(23))/(6.023xx10^(23))="0.25 mole of "CO_(2)` |
|
| 7102. |
Calculate the number of moles in each of the following. (i) 392 g of sulphuric acid (ii) 44.8 litres of sulphur dioxide at N.T.P. (iii) 6.022xx10^(22)" moles of oxygen" (iv) 8g of calcium |
|
Answer» SOLUTION :(i) 392 g of sulphuric acid Molar mass of `H_(2)SO_(4)=2xx1+32+4xx16=98g` `"98 g of sulphuric acid = 1 mole"` 392 g of sulphuric acid = 1 mol `xx(392)/((98g))="4 mol"` (II) 44.8 litres of sulphur dioxide at N.T.P 22.4 litres of sulphur dioxide at N.T.T. = 1 mol 44.8 litres of sulphur dioxide at N.T.P. `=("1 mole")/("(22.4 L)")xx(44.8L)="2.0 mol"` (III) `6.022xx10^(22)" molecules of oxygen"` `6.022xx10^(22)" molecules of oxygen = 1 mol"` `6.022xx10^(22)" molecules of oxygen = 1 mol "xx(6.022xx10^(22))/(6.022xx10^(23))="0.1 mol"` (iv) 8g oc CALCIUM Gram atomic mass of `Ca=40g` `"40 g of calcium= 1 mol"` `"8.0 g of calcium = 1 mol"xx((8.0g))/((40g))="0.2 mol."` |
|
| 7103. |
Calculate the number of moles in 24.092xx10^(22) molecules of water. |
|
Answer» SOLUTION :`"NUMBER of moles"=("No. of MOLECULES")/("Avogadro number")` `=(24.092xx10^(2))/(6.023xx10^(23))` `=4xx10^(22)xx10^(-23)=4xx10^(-1)` `="0.4 mole."` |
|
| 7104. |
Calculate the number of moles in 12.046xx10^(22) atoms of copper. |
|
Answer» SOLUTION :`"No. of moles of ATOMS"=("No. of atoms")/("Avogadro number")` `=(12.046xx10^(22))/(6.023xx10^(23))=2XX10^(-1)` `=0.2" mole."` |
|
| 7105. |
Calculate the number of moles in 46 g of sodium. |
|
Answer» Solution :`"Number of moles"=("MASS of the ELEMENT")/("ATOMIC mass of the element")` `=46//23` `="2 moles of SODIUM"` |
|
| 7106. |
Calculate the number of moles for a substance containing 3.0115xx10^(23) molecules in it. |
|
Answer» SOLUTION :`"NUMBER of moles "=("No. of molecules")/("Avogadro number")` `=(12.046xx10^(22))/(6.023xx10^(23))="0.5 MOLE"` |
|
| 7107. |
Calculate the number of moles :7.85 g of Fe |
|
Answer» |
|
| 7108. |
calculate thenumberof moleculespresentin the 36 gwater . |
|
Answer» Solution :The moleculesweightof `H_(2) O=18` Thus,18Gramsof ` H_(2)O`= 1moleof `H_(2)O` `36"gramsof"H_(2) O = 1 " 2moleof " H_(2)O` 1moleof `H_(2) O` `=6.023 xx 10^(23) ` MOLECULESOF `H_2O`. `=1.204xx 10 ^(24)`moleculesof ` H_2 O`. |
|
| 7109. |
Calculate the number of moles :5.0 L of 0.75 M Na_(2)CO_(3) |
|
Answer» |
|
| 7110. |
Calculate the number of moles :34.2 g of sucrose (C_(12)H_(22)O_(11)) |
|
Answer» |
|
| 7111. |
Calculate the number of molecules present in 22.0 g of CO_(2). |
|
Answer» |
|
| 7112. |
Calculate the number of molecules of 11 g of CO_(2). |
|
Answer» Solution :`"Gram molecular mass of "CO_(2)=44G` `"NUMBER of molecules"=("Avogadro number"XX"given mass")/("Gram molecular mass")` `=(6.023xx10^(23)xx11)/(44)` `=1.51xx10^(23)CO_(2)" molecules"` |
|
| 7113. |
Calculate the number of molecules in 54 gm of H_(2)O |
|
Answer» Solution :`"Number of molecules"=("AVOGADRO number"xx"Given mass")/("Gram MOLECULAR mass")` `"Number of molecules of water "=6.023xx10^(23)xx54//18` `=18.069xx10^(23)" molecules"` |
|
| 7114. |
Calculate the number of molecules in 360g of flucose. |
|
Answer» Solution :`"Number of molecules"=("AVOGADRO number"xx"GIVEN mass")/("GRAM molecular mass")` Gram molar mass of GLUCOSE `(C_(6)H_(12)O_(6))` `=(6xx12)+(12xx1)+(6xx16)` `=72+12+96=180g` `"Number of molecules"=(6.023xx10^(23)xx360)/(180)=6..023xx10^(23)xx2` `=12.046xx10^(23)" molecules (or)1.2046 "xx10^(24)" glucose molecules"` |
|
| 7115. |
Calculate the number of molecules in 11.2 litre of CO_(2) at S.T.P |
|
Answer» SOLUTION :`"NUMBER of MOLES of "CP_+(2)=("Volume at S.T.P")/("Molar volume")` ? `=11.2//22.4` `="0.5 MOLE"` `"Number of molecules of "CO_(2)="Number of molesof "CO_(2)xx"Avogadro.s number"` `=0.5xx6.023xx10^(23)` `=3.011xx10^(23)" molecules of "CO_(2)` |
|
| 7116. |
Calculate the number of molecule present in16.8 L of gas 'X' at STP. Also determine itsgram molecular weight if theabove sample weighs 26.625 g. |
|
Answer» SOLUTION :Volume of the given gas= 16.8 L at STP Number of moles present in that volume = 16.8/22.4 = 0.75 moles Number of MOLECULES in 0.75 moles = `0.75 xx 6.023 xx 10^(23) = 4.52 xx 10^(23)` Weight of 0.75 moles of the gasX = 26.625 g Gram MOLECULAR weight of 'X' = Weight /No. of moles = 26.625/0.75 = 35.5g |
|
| 7117. |
Calculate the number of kJ necessary to raise the temperature of 60g of aluminium from 35 to 55^(@)C. Molar heat capacity of Al is 24 J mol^(-1)K^(-1). |
|
Answer» |
|
| 7118. |
Calculate the number of atoms present in 1 gram of gold ("Atomic mass of Au = 198") |
|
Answer» Solution :`"NUMBER of ATOMS of Au"=("Mass of Au"XX"Avogadro.s number")/("ATOMIC mass of Au")` `"Number of atoms of Au"=(1)/(198)xx6.023xx10^(23)` `"Number of atoms of Auy"=3.042xx10^(21)g` |
|
| 7119. |
Calculate the number of atoms of oxygen and carbon in 5 moles of CO_(2). |
|
Answer» Solution :`"1 mole of "CO_(2)" contains 2 moles of OXYGEN"` `"5 moles of "CO_(2)" contain 10 moles of oxygen"` `"Number of atoms of oxygen"="Number of moles of oxygen"XX"Avogadro.s number"` `=10xx6.023xx10^(23)` `=6.023xx10^(24)" atoms of Oxygen"` `"1 mole of "CO_(2)" contains 1 mole of carbon"` `"5 moles of "CO_(2)" contains 5 moles of carbon"` `"No. of atoms of carbon"="NUMBR of moles of oxygen"xx"Avogadro.s number"` `=106.023xx10^(23)` `=6.023xx10^(24)"atoms of Oxygen"` `"1 mole of "CO_(2)" contains 1 mole of carbon"` `"5 moles of "CO_(2)" contains 5 moles of carbon"` `"No. of atoms of carbon"="No. of moles of carbon"xx"Avogadro.s number"` `=5xx6.023xx10^(23)` `=3.011xx10^(23)" atoms of Carbon"` |
|
| 7120. |
Calculate the number of atoms in each of the following: (i) 52 moles of He (ii) 52 u of He (iii) 52 g of He |
|
Answer» SOLUTION :(i) `"1 mol of He"=52xx6.022xx10^(23)" atoms"=3.131xx10^(25)" atoms"` (II) `"1 atom of He = 4 u of He"` `"4 u of He"="4u of He"` `therefore" 52 u of He"=(1)/(4)XX"52 atoms"="13 atoms"` (III) `"1 mole of He "=4g=6.022xx10^(23)" atoms"` `therefore"52 g of He"=(6.022xx10^(23))/(4)xx"52 atoms"=7.8286xx10^(24)" atoms"` |
|
| 7122. |
Calculate the number of atoms in each of the following : 52g of He. |
|
Answer» |
|
| 7123. |
Calculate the number of atoms in each of the following : 52 moles of Ar. |
|
Answer» |
|
| 7125. |
Calculate the mole fraction of the solute in a 1.00 molal aqueoussolution. |
|
Answer» |
|
| 7126. |
Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution. |
|
Answer» |
|
| 7127. |
Calculate themolarity of a solution obtained by mixing 250 mL of 0.5 MHClwith 750mL of 2 M HCl. |
|
Answer» `1.8` ` = ((0.5)250+(2xx750))/(250+750) = 1.625`N `:. ` Molarity = 1.625N(as HCl is MONOBASIC acid) |
|
| 7128. |
Calculate the maximum work obtained when 0.75 mole of an ideal gas expands isothermally and reversibly at 27^(@)C From a volume of 15L to 25L. |
|
Answer» |
|
| 7129. |
Calculate the mass of the following: 5xx10^(23) molecules of glucose |
|
Answer» SOLUTION :`"Molecular MASS of glucose"=180` `"Mass of glucose"=("Molecular mass"XX"number of particles")/("Avogadro.s number")` `=(180xx5xx10^(23))//6.023xx10^(23)` `=149.43g` |
|
| 7130. |
Calculate the mass of the following: 2.23 litre of SO_(2) gas at S.T.P |
|
Answer» Solution :`"Molecular MASS of "SO_(2)=32+(16xx2)` `=32+32=64` `"Number of moles of "SO_(2)=("Given volume of "SO_(2)" at S.T.P")/("Molar volume "SO_(2)"at S.T.P")` `"Number of moles of "SO_(2)=(2.24)/(2.24)="0.1 MOLE"` `"Number of moles"=("Mass")/("Molecular mass")` `"Mass"="No. of moles"XX"molecular mass"` `"Mass"=0.1xx64` `"Mass of "SO_(2)=6.4g` |
|
| 7131. |
Calculate the mass of the following: 1//51xx10^(23) molecules of water |
|
Answer» Solution :`"MOLECULAR mass of "H_(2)O=18` `"NUMBER of moles"=("Number of MOLECULES of water")/("Avogadro.s number")` `=1.51xx10^(23)//6.023xx10^(23)` `=1//4` `="0.25 moles"` |
|
| 7132. |
Calculate the mass of the following: 0.3 mole of aluminium (Atomic mass of Al = 27) |
|
Answer» Solution :`"Number of MOLES"=("Mass of AL")/("ATOMIC mass of Al")` `"Mass"="No. of moles"XX"atomic mass"` `"So, MAS of Al"=0.3x23` `=8.1g` |
|
| 7133. |
Calculate the mass of glucose in 2xx10^(24) molecules. |
|
Answer» Solution :`"Gram molecular MASS ofglucose = 180 g"` `"Mass of glucose"=(180xx2xx10^(24))/(6.023xx10^(23))` `=597.7g` |
|
| 7134. |
Calculate the mass of aluminium oxide which contains double the number of oxygen atoms in 192g of oxygen gas. |
|
Answer» SOLUTION :(i) atomicity of oxygen (ii) CHEMICAL composition of `Al_(2)O_(3)` (iii) determination of mass of `Al_(2)O_(3)` (iv) 408g |
|
| 7135. |
Calculate the mass of 18.069xx10^(23) molecules of SO_(2) |
|
Answer» Solution :`"Mass of the SUBSTANCE"=("Gram molecular mass"XX"NUMBER of particles")/("AVOGADRO number")` Gram molecular mass of `SO_(2)=32+16(2)=64G` `"Mass of "SO_(2)=(64xx18.069xx10^(23))/(6.023xx10^(23))` `=64xx3=192g` |
|
| 7136. |
Calculate the mass of 12.046xx10^(23) molecules of CaO. |
|
Answer» SOLUTION :`"GRAM MOLECULAR mass of "CaO=40+16=56g` `"Mass of CaO"=(56xx12.046xx10^(23))/(6.023xx10^(23))=56xx2` `=112g` |
|
| 7137. |
Calculate the mass of 0.5moles of O_(2) gas |
|
Answer» 16g |
|
| 7138. |
Calculate the mass of 0.5 mole of iron. |
|
Answer» SOLUTION :`"Mass"="Atomic mass"xx"NUMBER of MOLES"` `"Mass of iron"=55.9xx0.5=27.95g` |
|
| 7139. |
Calculatethe grammolecularmass of thefollowingCO_(2) |
|
Answer» SOLUTION :`CO_(2)` Atomicmassesof C =12 , O=16 Grammolarmassof `CO_(2)` `= (12xx1 )+(16xx2 )` `=12 +32` Grammolecularmassof `CO_(2)= 44 g` |
|
| 7140. |
Calculate the gram molar mass of the following. (i) H_(2)O (ii) CO_(2) (iii) Ca_(3)(PO_(4))_(2) |
|
Answer» Solution :(i) `H_(2)O` `"Atomic MASSES of H"=1, O=16` `"Gram molar MASS of "H_(2)O=(1xx2)+(16xx1)` `=2+16` `"Gram molar mass of "H_(2)O=18g` (ii) `CO_(2)` `"Atomic masses of C"=12, O=16` `"Gram molar mass of "CO_(2)=(12xx1)+(16xx2)` `=12+32` `"Gram molar mass of "CO_(2)=44g` (iii) `Ca_(3)(PO_(4))_(2)` `"Atomic masses of CA"=40, P=30, O=16`. `"Gram molar mass of "Ca_(3)(PO_(4))_(2)=(40xx3)+[30+(16xx4)]xx2` `=120+(94xx2)` `=120+188` `"Gram molar mass of "Ca_(3)(PO_(4))_(2)=308g` |
|
| 7141. |
Calculate the entropy change in the surroundings when 1.00 mol of H_(2)O(I) is formed under standard conditions, Delta_(f) H^(theta) = - 286 kJ mol^(-1) |
| Answer» SOLUTION :`959.7 JK^(-1) MOL^(-1)` | |
| 7142. |
Calculate the enthalpy change for the reaction: H_(2) (g) + Br_(2) (g) rarr 2HBr(g) Given the bond enthalpies H_(2), Br_(2) and HBr are 435 kJ mol^(-1), 192 kJ mol^(-1) and 368 kJ mol^(-1) respectively. |
| Answer» SOLUTION :`-109 KJ MOL^(-1)` | |
| 7143. |
Calculate the difference in the formulae and molecular masses for (a) CH_(3)OH and C_(2)H_(5)OH, (b) C_(2)H_(5)OH and C_(3)H_(7)OH and ( c ) C_(3)H_(7)OH and C_(4)H_(9)OH |
Answer» SOLUTION :
|
|
| 7144. |
Calculate the average atomic mass of carbon, if the natural abundance of C-12 and C-13 are 98.90% and 1.10% respectively. |
|
Answer» Solution :AVERAGE ATOMIC mass of CARBON `=(12xx(98.9)/(100))+(13xx(1.1)/(100))` `=(12xx0.989)+(13xx0.011)` `=11.868+0.143="12.011 amu"` |
|
| 7145. |
Calculate the amount of water (g) produced by the combustion of 16 g of methane |
|
Answer» |
|
| 7146. |
Calculate the amount of sodium oxide formed when 2.3 g of sodium reacts with 3.2 g of oxygen. |
|
Answer» Solution :`{:(4Na+O_(2) to 2Na_(2)O),("4 moles" " 1 MOLE" " 2 moles"):}` `4 xx 23` g of sodium reacts with 32 g of `O_(2)` 2.3g of sodium reacts with =`(2.3 xx 32)/(4 xx 23) = 0.8` g As the amount of oxygen PRESENT is 3.2 g, all the oxygen is not utilised for FORMATION of sodium oxide. The amount of sodium oxide formed is determined by sodium.`4 xx 23` g of sodium PRODUCES `2 xx 62` g of sodium oxide. 2.3 g of sodium produces = `(2.3 xx 2 xx62)/(4 xx 23) = 3.1` g Amount of sodium oxide formed = 3.1 g. |
|
| 7147. |
Calculate % of S in H_(2)SO_(4) |
|
Answer» Solution :`"Molar mass of "H_(2)SO_(4)=(1xx2)+(32xx1)+(16xx4)` `=2+32+64` = 98g `"% of S in "H_(2)SO_(4)=("Mass of sulphur")/("Molar mass of "H_(2)SO_(4))=100` `"% of S in "H_(2)SO_(4)=(32)/(98)XX100` `=32.65%` |
|
| 7149. |
Calculate molarity of water if its density is 1.00 g mL^(-1). |
|
Answer» |
|