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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.
| 151. |
Assertion(A) One atomic mass unit is defined as one twelth of the mass of one carbon-12 atom. Reason(R) Carbon-12 isotope is the most abundant isotope of carbon and has been chosen as standard.A. Both A and R are true and R is the correct explnanation of A.B. Both A and R true but R is not the correct explanation of A.C. A is true but R is false.D. Both A and R are false. |
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Answer» Both Assertion and Reason are true but Reason is not the correct explanation of Assertion. Atomic masses of the elements obtained by scientists by comparing with the mass of carbon comes out to be close to whole number value. |
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| 152. |
Match the statements of column I with values of Column II `{:(ulbar(" ""Column I"" ""Column II"" ")),(ul(A." ""Different number of atoms"" "p." "4.25g NH_(3) and 4.5 g of H_(2)O)),(ul(B." ""Same number of molecules" " "q." "2.20 g CO_(2) and 0.90 g H_(2)O)),(ul(C." ""Same numbers of atoms as well as molecules"" "r." "4.0 g CH_(3)Cl and 5.0 g NH_(3))),(ul(D." ""Different numbers of atoms as well as molecules"" "s." "4.80 g O_(2) and 2.82 g CO)):}` |
| Answer» Correct Answer - `A rarr s; B rarr p; C rarr q; D rarr r` | |
| 153. |
Assertion(A) One atomic mass unit is defined as one twelth of the mass of one carbon-12 atom. Reason(R) Carbon-12 isotope is the most abundant isotope of carbon and has been chosen as standard.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Choosing C-12 as the standard, the relative masses of atoms come out to be close to whole number values and ambiguity of oxygen as reference was solved. | |
| 154. |
On a conventional scale, atomic weight of sulphur is 32. If on a new scale, an amu is defined as one atom of C-12 isotope, what would be the atomic weight of sulphur on this new scale? |
| Answer» Correct Answer - 8 | |
| 155. |
If equal volumes of `3.5 M CaCl_(2)` and `3.0 M NaCl` are mixed, what would be the molarity of chloride ion in the final soluton? |
| Answer» Correct Answer - 5 | |
| 156. |
The molar heat capacity of water at constant pressure, C, is `75 JK^(-1) mol^(-1)`. When 1.0 kJ of heat is supplied to 100 g water which is free to expand, the increase in temperature of water is :A. 6.6 KB. 1.2 KC. 2.4 KD. 4.8 K |
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Answer» Correct Answer - C Heat capacity of water per gram `=75/18=4.17` `Q=mST` `1000=100xx4.17xxT` `T=1000(100xx4.17)=2.4 K`. |
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| 157. |
Washing soda `(Na_(2)CO_(3).10 H_(2)O)` is widely used in softening of hard water. If 1 L of hard water requires 0.0286 g of washing soda, the hardness of `NaOH` in ppm is |
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Answer» Correct Answer - 4 Mw of washing soda `=286`, Mw of `NaOH=40" g mol"^(-1)` `Ca(OH)_(2)+Na_(2)CO_(3) rarr CaCO_(3)+2 NaOH` `0.0286 g Na_(2)CO_(3). 10 H_(2)O =0.0286/(286//2)xx10^(3)=0.2` meq 0.2 meq of `Na_(2)CO_(3)=0.2/2` mmol of `NaOH` in 1L `=(0.1xx10^(-3))xx40g` in `10^(3)` mL. `=0.01 g` in `10^(3)` mL ppm `=(4xx10^(-3)xx10^(6))/10^(3)=4` ppm. |
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| 158. |
A bivalent metal has an equivalent mass of 32. The molecular mass of the metal nitrate isA. `168`B. `192`C. `188`D. `182` |
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Answer» Correct Answer - C Atomic mass of metal = Eq. mass `xx` valency `=32xx2=64` Formula of metal nitrate `= M(NO_(3))_(2)` `:.` Molecular mass of the metal nitrate `= 64+2(14+3xx16)=188` |
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| 159. |
Equivalent weight of a bivalent metal is `37.2`. The molecular weight of its chloride isA. (a)`412.2`B. (b)`216`C. (c )`145.4`D. (d)`108.2` |
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Answer» Correct Answer - C Equivalent weight of bivalent metal`=37.2` `:. ` Atomic weight of metal`=37.2xx2=74.4` `:.` Formula of chlorid`=MCl_(2)` Hence, molecular weight of chloride `(MCl_(2))=47.4+2xx35.5=145.4` |
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| 160. |
A bivalent metal has an equivalent mass of 32. The molecular mass of the metal nitrate isA. 168B. 192C. 188D. 182 |
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Answer» Correct Answer - C Given, equivalent mass of bivalent metal, `M^(2+)=32` `:.` Atomic mass of `M=32xx2=64` The metal nitrate formed has the formula `M(NO_(3))_(2)` `:.` Molecular mass of the metal nitrate `=64+28+96=188`. |
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| 161. |
The molecular weight of a gas is `45`. Its density at STP isA. 22.4B. 11.2C. 5.7D. `2.0` |
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Answer» Correct Answer - D The density of gas `=("molecular wt. of metal")/("volume")` `=45/22.4=2" gm litre"^(-1)` |
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| 162. |
A compound possesses 8% sulphur by mass. The least molecular mass isA. `200`B. `400`C. `155`D. `355` |
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Answer» Correct Answer - B A compound has sulphur = 8% `:.` 8g of S is present in 100 g `:.` 32 g of S is present in `(100)/(8)xx32=400g` (Least molecular mass is when the molecule is supposed to contain one atom of sulphur) |
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| 163. |
Which of the following has least massA. 2 g atom of nitrogenB. `3xx10^(23)` atom of CC. 1 mole of SD. 7.0 g of Ag |
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Answer» Correct Answer - B (a) `2` gm atom of nitrogen `=28 gm` (b) `6xx10^(23)` atoms of C has mass `=12` gm `3xx10^(23)` atoms of C has mass `=(12xx3xx10^(23))/(6xx10^(23))=6 gm` (c) 1 mole of S has mass `=32` gm (d) 7.0 gm of Ag So, lowest mass `=6` gm of C. |
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| 164. |
A compound possesses `8%` sulphur by mass. The least molecular mass is?A. (a)`200`B. (b)`400`C. (c )`155`D. (d)`355` |
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Answer» Correct Answer - c `:. 8 g` sulphur is present in `100 g` of substance `:. 32 g` sulphur will present `=100/8xx32=400`. |
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| 165. |
The one which has least mass, isA. 2 g atom of NB. `3 xx 10^(23)` atoms of CC. 1 mole of SD. 7.0 g of Ag |
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Answer» `2g` atom of `N_(2)=2xx14=28g` `3xx10^(23)C` atoms `=(3xx10^(23))/(6xx10^(23))xx12=6g` 1mole of S= 32g |
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| 166. |
Which of the following contains maximum number of atom?A. `2*0` mol of `S_(8)`B. `6*0` mol of SC. `5*5` mol of `SO_(2)`D. `44*8 L` of `CO_(2)` at S.T.P. |
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Answer» Correct Answer - C (A) Atoms in `2*0` mol of `S_(8) = 2xx8xx6*02xx10^(23) = 9*632xx10^(24)` (B) Atoms in `6*0` mol of S `=6xx6*02xx10^(23)=3*612xx10^(24)` (C) Atoms in `5*5` mol of `SO_(2) = 3xx5*5xx6*02xx10^(23) = 9*93xx10^(24)` (D) Atoms in `4*48L` of `CO_(2)` at S.T.P. `= (3xx4*48xx6*02xx10^(23))/(22*4)=4*816xx10^(23)` |
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| 167. |
A person adds 1.71 g of sugar `(C_(12)H_(22)O_(11))` in order to sweeten his tea. The number of carbon atoms added are (molecular mass of sugar = 342)A. `3.6 xx 10^(22)`B. `7.2 xx 10^(21)`C. 0.05D. `6.6 xx 10^(22)` |
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Answer» Number of moles of sugar `=(1.71)/(342)=0.005` `because` 1 mole sugar contains `=12xx6.02xx10^(23)` `therefore 0.005` mole of sugar will contain `=12xx0.005xx6.02xx10^(23)` `=0.36xx10^(23)=3.6xx10^(22)` C-atoms |
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| 168. |
A person adds `1*71` gram of sugar `(C_(12)H_(22)O_(11))` in order to sweeten his tea. The number of carbon atoms added are (mol. mass of sugar - 342)A. `3*6xx10^(22)`B. `7*2xx10^(21)`C. `0*05`D. `6*6xx10^(22)` |
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Answer» Correct Answer - A Moles of sugar added `= (1*71)/(342) = 5xx10^(-3)` Carbon atoms added `=12xx5xx10^(-3)xx6*02xx10^(23)` `= 3.61xx10^(22)` |
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| 169. |
2.82g of glucose is dissolved in 30g of water. The mole fraction of glucose in the solution isA. 0.01B. `0.99`C. `0.52`D. 1.66 |
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Answer» No. of moles of glucose `=(2.82)/(180)=0.01567` No. of moles of water `=30/18=1.667` Total of moles of solution =0.01567+1.667=1.683 Moles fraction glucose `=(0.01567)/(1.683)=0.0093=0.01` |
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| 170. |
Molarity of pure water isA. `18 M`B. `10 M`C. `55.5 M`D. `1000 M` |
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Answer» Correct Answer - C Molarity of pure water implies the number of moles of `H_(2) O` per liter of water. Since the density of `H_(2) O` is `1g mL^(-1)`, the mass of `1L (1000 mL)` of water will be `1000g`. Thus, the number of moles of `H_(2) O` in `1000g` will be `n_(H_(2)O) = ("Mass of" H_(2) O)/("Molar Mass of" H_(2) O) = (1000g)/(18.0g mol^(-1)) = 55.5 mol` As these moles are present in `1L` of `H_(2) O`, we say that molarity `(mol L^(-1))` of `H_(2)O` is `55.5M`. |
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| 171. |
Molarity of pure `D_(2)O` will beA. `55.56`B. 1C. 50D. 10 |
| Answer» Correct Answer - C | |
| 172. |
A sample of pure calcium weighing 1.35 g was quantitatively converted to 1.88 g of pure CaO . Atomic mass of calcium isA. 20B. 40C. 60D. 30 |
| Answer» Correct Answer - B | |
| 173. |
Which of the following is a pure substance?A. GasolineB. Distilled waterC. Iodized table saltD. Liquefied petroleum gas |
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Answer» Correct Answer - B Others are mixtures |
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| 174. |
To neutralise 20 ml of M/10 sodium hydroxide, the volume of M/20 hydrochloric acid required isA. (a)10 mlB. (b)15 mlC. (c )20 mlD. (d)40 ml |
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Answer» Correct Answer - D `NaOH" " HCl` `N_(1)V_(1)=N_(2)V_(2), 20xx1/10=1/20xxV,V=40 ml` |
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| 175. |
When 100 ml of 1 M NaOH solution and 10 ml of 10 N `H_(2)SO_(4)` solution are mixed together, the resulting solution will beA. AlkalineB. AcidicC. Atrongly acidicD. Neutral |
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Answer» Correct Answer - D For `NaOH, M=N` `N_(1)V_(1)=100 ml xx 1N=100` ml (N) For `H_(2)SO_(4), N_(2)V_(2)=10mlxx10 N=100` ml (N) Hence, `N_(1)V_(1)=N_(2)V_(2)`. |
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| 176. |
What is the volutme of `CO_(2)` liberted in litres at 1 atmosphere and `0^(@)C` when 10% of 100 pure calcium carbonate is treated with excess dilute sulphuric acid? (at mass of Ca=40, C=12, O=16)A. `0.224`B. `2.24`C. `22.4`D. 224 |
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Answer» Correct Answer - B `{:(CaCO_(3)+H_(2)SO_(4) rarr CaSO_(4)+H_(2)O+CO_(2)),(" 1 mol 1 mol"),(" 100 g 22.4 litre"),(" 10 g 2.24 litre"):}` |
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| 177. |
At what temperature the Celsius and Fahrement readings have the same numerical value?A. `-35^(@)`B. `-40^(@)`C. `-45^(@)`D. `-30^(@)` |
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Answer» Correct Answer - B `.^(@)C = (.^(@)F - 32)/(1.8) implies X = (X - 32)/(1.8) implies X = -40` |
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| 178. |
The number of significant figures in pi `(pi)` isA. threeB. oneC. infiniteD. two |
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Answer» Correct Answer - C `pi = 22//7`, both numerator and denominator are exact numbers with infinite number of significant figures. |
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| 179. |
Hydrolic acid solution A and B have concentration of 0.5 N and 0.1N respectively. The volume of solutions A and B required to make 2 litres of 0.2 N hydrochloric areA. (a)`0.5 L of A+1.5L of B`B. (b)`1.5 L of A+0.5L of B`C. (c )`1.0 L of A+1.0L of B`D. (d)`0.75 L of A+1.25L of B` |
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Answer» Correct Answer - A `NV=N_(1)V_(1)+N_(2)V_(2)` `0.2xx2=0.5x+0.1(2-x)` `0.4=0.5x=0.2-0.1x` `0.0=0.4x` `x=1/2L=0.5L` |
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| 180. |
Assertion(A) The empirical mass of ethene is half of its molecular mass. Reason(R) The empirical formula represents the simplest whole number ratio of various atoms present in a compound.A. Both A and R are true and R is the correct explnanation of A.B. A is true but R is false.C. A is false but R is true.D. Both A and R are flase. |
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Answer» Both assertion and Reason are true and Reason is the correct explanation of Assertion. The molecualr formula of ethene is `C_(2)H_(4)` and itsw empirical formula is `CH_(2)` Thus, Molecular formula =Empirical formula `xx2` |
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| 181. |
Which of the following fills any container completely irrespective of the amount?A. GasB. LiquidC. SolidD. Both(1) and (2) |
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Answer» Correct Answer - A A gas tends to fill up all the available space as the intermolecular attractions are weak. |
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| 182. |
How many millilitres of `0.1 NH_(2)SO_(4)` solution will be required for complete reaction with a solution containing 0.125 g of pure `Na_(2)CO_(3)`?A. (a)23.6 mLB. (b)25.6 mLC. (c )26.3 mLD. (d)32.6 mL |
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Answer» Correct Answer - A Meq of `H_(2)SO_(4)`=Meq of `Na_(2)CO_(3)` `0.1xxV/1000=0.125/106xx2` `V=23.6 mL` |
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| 183. |
Composition fo a colorless liquid is `84.1% C` and `15.9% H` by mass. Its empirical formula isA. `C_(3) H_(7)`B. `C_(4) H_(9)`C. `C_(5) H_(11)`D. `CH_(3)` |
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Answer» Correct Answer - B `n_(C) = (84.1g)/(12.01g mol^(-1)) = 7.00 mol C` `n_(H) = (15.9g)/(1.008g mol^(-1)) = 15.8 mol H` Dividing by the smaller interger 4, we get `n_(C) : n_(H) = (7.00)/(7.00) : (15.8)/(7.00) = 1: 2 26` Mutiplying by the smallest interger 4, we get `(1xx4) : (2.26xx4) = C_(4) H_(9)` |
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| 184. |
The mass of oxalic acid crystals `(H_(2)C_(2)O_(4).2H_(2)O)` required to prepare 50 mL of a 0.2 N solution is:A. (a)4.5 gB. (b)6.3 gC. (c )0.63 gD. (d)0.45 g |
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Answer» Correct Answer - C `H_(2)C_(2)O_(4). 2H_(2)O=2+24+64+36=126` and Equivalent wt. `=[126/2]` `0.2=(Wxx1000)/((126/2)xx50) :. W=0.63 g` |
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| 185. |
Two samples of lead oxide were separately reduced to metallic lead by heating in a current of hydrogen. The weight of lead from one oxide was half the weight of lead obtained from the other oxide. The data illustratesA. (a)Law of reciprocal proportionsB. (b)Law of constant proportionsC. (c )Law of multiple proportionsD. (d)Law of equivalent proportions |
| Answer» Correct Answer - C | |
| 186. |
4g of copper was dissolved in conc. Nitric acid. The copper nitrate so obtained on strong heating gave 5 g of its oxide. The equivalent weight of copper isA. `23`B. `32`C. `12`D. `20` |
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Answer» Correct Answer - B 4g of copper gives 5 g of its oxide `:.` 4 g of copper combines with 1 g of oxygen `:.` Mass of copper that combines with 8g of oxygen `8xx4=32` `:.` Eq. mass of copper = 32 |
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| 187. |
2,75 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper that remained was 2.196 g. Another experiment, 2.358 g of copper was dissolved in nitric acid and the resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed was 2.952 g. Show that these results illustrate law of constant composition. |
| Answer» In both cases percentage of oxygen is the same, so the law of constant composition is illustrated. | |
| 188. |
`1.59 g` of first sample fo cupric oxide `(CuO)` on comple reduction by hydrogen `(H_(2))` gas gave `1.27 g` of pure copper `(Cu)` metal. Secound pure sample of curpic oxide weighing `3.18 g` yieled `2.54 g` of pure copper metal on complete refuction by hydrogen gas. Show that the law of definite proportions is valid. Strategy: Find the ratio by mass of copper to oxygen in both the samples. |
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Answer» Pure sample `-I` `m_(CuO) = 1.59 g` `m_(Cu) = 1.27 g` `:. M_(O) = (1.59 g) - (1.27 g) = 0.32 g` Now, `m_(cu) : m_(O) = 1.27g : 0.312 g = ((1.27)/(0.32)): ((0.32)/(0.32)) = 3.9 : 1` Pure sample - `II` `m_(CuO) = 3.18 g` `m_(Cu) = 2.54 g` `:. m_(O) = (3.18 g) - (2.54 g) = 0.64 g` Now, `m_(Cu) : m_(O) = (2.54 g): (0.64 g) = ((2.54)/(0.64)) : ((0.64)/(0.64))` As both the pure sample contain the samples contan the same ratio by mass of copper to oxygen, the law of definite proptions is valid. |
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| 189. |
0.75 g platinic chloride, a monoacid base on ignition gave 0.245 g playinum. The molecular weight of the base isA. `75.0`B. `93.5`C. `100`D. `80.0` |
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Answer» Correct Answer - B `("Wt. of "B_(2)H_(2)PtCl_(6))/(2B+410)=("Wt. of Pt")/195` `:. 0.75/(2B+410)=0.245/195 implies B=93.5` Eq. wt. of base `=93.5`, since it is monoacidic. `:.` Mol. Wt. of base `=93.5xx1=93.5` |
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| 190. |
Copper oxide was prepared by two different methods. In case, 1.75 g of the metal gave 2. 19 g of oxide. In the second case, 1.14 g of the metal gave 1.43 g of the oxide, show that the given data illustrate the law of constant proportions. |
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Answer» In case I, mass of copper = 1.75 g Mass of copper oxide = 2.19 g % of copper in the oxide `= ("Mass of copper")/("Mass of copper oxide")xx 100` `=(1.75)/(2.19)xx100 = 79.9 %` `therefore %` of oxygen `= 100 - 79.9 = 20.1 %` In case II, mass of copper = 1.14 g Mass of copper oxide = 1.43 g % of copper in the oxide `=(1.14)/(1.43)xx100 = 79.7 %` % of oxygen = 100 - 79.7 = 20.3 %. Thus, copper oxide prepared by any of the given methods contain copper and oxygen in the same proportion by mass (with the experimental error). Hence, it proves the law of constant proportions. |
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| 191. |
Hydrogen and oxygen are known to from two compounds. The hydrogen content in one one of these is 5.93% while in the other it is 11.2%. Show that this data illustrates the law of multiple proportions. |
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Answer» In the first compound Hydrogen = 5.93 % Oxygen = `(100-5.93)% = 94.07 %` In the second compound Hydrogen = 11.2 % Oxygen `= (100-11.2)%=88.8%` In the first compound the number of parts by mass of oxygen that combine with one part by mass of hydrogen `=(94.07)/(5.93)=15.86` parts In the second compound the number of part by mass of oxygen that combine with one part by mass of hydrogen `=(88.8)/(11.2)=7.9` parts. The ratio of masses pf oxygen that combine with fixed mass (1 part) by mass of hydrogen is 15.86 : 7.9 or 2 : 1. Since this ratio is a simple whole number ratio, hence the given data illustrates the law of multiple proportions. |
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| 192. |
Around `20%` surface sites have adsorbed `N_(2)`. On heating `N_(2)` gas evolved form sites and were collected at 0.001 atm and 298 K in a container of volume `2.46cm^(3)` the density of surface sites is `6.023xx10^(14)cm^(-2)` and surface area is `1000cm^(2)` find out the number of surface sites occupied per molecule of `N_(2)`. |
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Answer» Correct Answer - -2 Partial ideal gas law : `pV = nRT` `n(N_(2)) = (pV)/(RT) = (0.001 xx 2.46)/(0.082 xx 298) = 10^(7)` `rArr` Number of molecules of `N_(2) = 6.023 xx 10^(23) xx 10^(-7)` `= 6.023 xx 10^(16)` Now, total surface sites available `= 6.023 xx 10^(14) xx 1000 = 6.023 xx 10^(17)` Surface sites used in adsorption `= (20)/(100) xx 6.023 xx 10^(17)` `= 2 xx 6.023 xx 10^(16)` `rArr` Sites occupied per molecules `= ("Number of sites")/("Number of molecules") = (2 xx 6.023 xx 10^(16))/(6.023 xx 10^(16)) = 2` |
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| 193. |
Among the following pairs of compounds, the one that illustrates the law of multiple proportions isA. (a)`NH_(3) and NCI_(3)`B. (b)`H_(2)S and SO_(2)`C. (c )`CuO and Cu_(2)O`D. (d)`CS_(2) and FeSO_(4)` |
| Answer» Correct Answer - C | |
| 194. |
An ore contains 1.34% of the mineral argentite, `Ag_(2)S`, by mass. How many gram of this ore would have to be processed in order to obtain 1.00 g of pure solid silver, Ag?A. `74.6 g`B. `85.7 g`C. `107.9 g`D. `134.0 g` |
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Answer» Correct Answer - C 100 g ore = 1.34 g of `Ag_(2)S` Molar mass of `Ag_(2)S = 248 mol^(-1)` 248 g of `Ag_(2)S = 2xx108g of Ag = 216 g of Ag 1.00 g of Ag` `= 1.00 g of Agxx((248AG_(2)S)/(216gAg))xx((100gore)/(1.34gAg_(2)S))` `= 85.68 g = 85.7 g` |
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| 195. |
A sample of pure carbon dioxide, irrespective of its source contains `27.27%` carbon and `72.73%` oxygen. The data supportA. (a)Law of constant compositionB. (b)Law of conservation of massC. (c )Law of reciprocal proportionsD. (d)Law of multiple proportions |
| Answer» Correct Answer - A | |
| 196. |
A sample of copper sulphate pentahydrate, `CuSO_(4). 5H_2O` contains `3.782g` of `Cu`. How many grams of oxygen are in this sample?A. `0.952 g`B. `3.809 g`C. `4.761 g `D. `8.576 g` |
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Answer» Correct Answer - D `CuCO_(4).5H_(2)O` has 1 mol of copper and 9 moles of oxygen atoms. `63.5 g Cu = 9xx16 g` of oxygen `3.782 g Cu = (9xx16xx3.782)/(63.5) = 8.576` of oxygen |
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| 197. |
A sample of ammonium phosphate, `(NH_(4))_(3)PO_(4)` contains `3.18` mol of hydrogen atoms. The number of moles of oxgen atoms in the sample is .A. `0.265`B. `0.795`C. `1.06`D. `3.18` |
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Answer» Correct Answer - C In one mole of `(NH_(4))_(3)PO_(4)`, 12 moles of hydrogen atoms 4 moles of oxygen atoms, 3 moles of nitrogen atoms and 1 mole of phosphorus atoms are present If 3.18 mole of hydrogen atom are there, number of moles of oxygen atoms `= (3.18)(3)mL = 1.06 mol` |
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| 198. |
`74.5 g` of a metallic chloride contain `35.5 g` of chlorine. The equivalent weight of the metal isA. (a)`19.5`B. (b)`35.5`C. (c )`39.0`D. (d)`78.0` |
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Answer» Correct Answer - C Weight of metallic chloride`=74.5` Weight of chlorine`=35.5` :. Weight of metal`=74.5-35.5=39` Equivalent weight of metal `=("weight of metal")/("weight of chlorine")xx35.5` `=39/35.5xx35.5=39` |
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| 199. |
`74.5 g` of a metallic chloride contain `35.5 g` of chlorine. The equivalent weight of the metal isA. 19.5B. 35.5C. `39.0`D. `78.0` |
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Answer» Correct Answer - C Wt. of metallic chloride =74.5 Wt. of chlorine `=35.5` `:.` wt. of metal `=74.5-35.5=39` Equivalent weight of metal `=("weight of metal")/("weight of chlorine")xx35.5` `=39/35.5xx35.5 =39` |
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| 200. |
A sample of ammonium phosphate `(NH_(4))_(3) PO_(4)` contains 3.18 moles of hyrogen atoms . The number of moles of oxygen atoms in the sample isA. `0.265`B. `0.795`C. `1.06`D. `4.00` |
| Answer» Correct Answer - C | |