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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.
| 51901. |
0.96 g of HI were, heated to attain equilibrium 2HI hArr H_(2)+I_(2). The reaction mixture on titration requires 15.7 mL of N//10 hypo solution. Calculate the degree of dissociation of HI. |
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Answer» mEq of `I_(2)` at equilibrium =mEq of hypo used for reaction mixture. `w_(I_(2))/Exx1000=15.7xx1/10` `:. (w/E)` of `I_(2)=1.57xx10^(-3)` `:.` moles of `I_(2)` formed `=(1.57xx10^(-3))/2=0.785xx10^(-3)` `x/2=0.785xx10^(-3) rArr x=1.57xx10^(-3)` `:.` Degree of dissociation of HI `("or " alpha_(HI))` `=("moles dissociated")/("moles TAKEN")=(1.57xx10^(-3))/(7.5xx10^(-3))` `alpha_(HI)=0.209` or `20.9%` |
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| 51902. |
0.9367 g of cadmium combine with chlorine to form 1.5276 g of CdCl_2- Find the equivalent mass of cadmium. |
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Answer» SOLUTION :Mass of cadmium taken = 0.9367 g Mass of `CdCl_2` formed = 1.5276 g Mass of chlorine that combines with 0.9367 g of cadmium = 1.5276 - 0.9367 = 0.5909 g HENCE, The EQUIVALENT weight of cadmium `=("Mass of the metal")/("Mass of chlorine") xx 35.5` `=0.9367/0.5909 xx 35.5 = 56.27` |
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| 51903. |
0.92 g of an organic compound was analysed by combustion method. The mass of the U-tube increased by 1.08 g. what is the percentage of hydrogen in the compound? |
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Answer» 0.1304 18g `H_(2)O-=2` g H `1.08" g "H_(2)O=(2)/(18)xx1.08` % of hydrogen=`(2)/(18)xx(1.08)/(0.92)xx100=13.04%` |
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| 51904. |
0.9031 g of a mixture of NaCI and KCI on treatment with H_2SO_4 yielded 1.0784 g of a mixture of Na_2SO_4 and K_2SO_4.Calculate the percent composition of the mixture. |
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Answer» `underset(2 xx 58.44 g)(2NaCl) + H_(2)SO_(4) to underset(142.04 g) + 2HCl` `underset(2 xx 74.55 g)(2KCl) + H_(2)SO_(4) to underset(174.26 g)(K_(2)SO_(4)) + 2HCl` `therefore 2 xx 58.44` of NaCl give `Na_(2)SO_(4) = 142.04 g` `therefore x g` of NaCl will give `Na_(2)SO_(4) =(142.04)/(2 xx 58.44) xx x` Similarly, the amount of `K_2So_4` formed by (0.9031 x) g of KCl `=(174.26)/(2 xx 74.55) xx (0.9031 -x)` `therefore` The amount of the resulting mixture of sulphates =1.0784 g `therefore 142.04/(2 xx 58.44 xx x) + [174.26/(2 xx 74.55) xx (0.9031 -x)] = 1.0784` which gives x=0.4829 g `therefore` Percentage of NaCl in the given mixture `=0.4829/0.9031 xx 100 = 53.47` and the percentage of KCI in the mixture `=100-53.47 = 46.53` |
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| 51905. |
0.900g of a solute was dissolved in 100 ml of benzene at 25^(@)C when its density is 0.879 g/ml. This solution boiled 0.250^(@)C higher than the boiling point of benzene. Molal elevation constant for benzene is "2.52 K.Kg.mol"^(-1). Calculate the molecular weight of the solute. |
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Answer» SOLUTION :`"Weight of benzene "=100xx0.879=87.9g` Molality of solution, m `=(0.900//M_(2))/(87.9)xx1000` `=(900)/(87.9M_(2))` `DeltaT_(b)=K_(b)m" (or)0.52"XX(900)/(87.9M_(2))` `therefore""M_(2)=(900xx2.52)/(87.9xx0.250)` `M_(2)="103.2g/mole"` `therefore""{:("Molecular weight"),("of the SOLUTE"):}}="103.2 g/mole"` |
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| 51906. |
0.84g iron are containing X per cent of iron was takenin a solution containg all the iron in ferrous state. The solution required X ml of a porassium dichromate soltuionfor oxidation of ironcontent to ferric state. Calculate the strength of potassiumdichromate solution. |
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| 51907. |
0.80 g of a substance was digested with sulphuric acid and then distilled with an excess of caustic soda. The ammonia gas evolved was passed through 100 ml of 1N H_(2)SO_(4). The excess of the acid required 80 ml of 1N caustic soda solution for its complete neutralisation. Calculate the percentage of nitrogen in the organic compound. |
| Answer» SOLUTION :`N = 35%` | |
| 51908. |
0.79 g of iron unite directly with 0.4 g sulphur to form ferrous sulphate. If 2.8 g of iron are dissolved in dilute HCl and excess of sodium sulphide solution is added, 4.4 g iron sulphite is precipitated. Show that the data is according to Law of constant composition. |
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Answer» Solution :In the first sample : Ratio by mass of IRON and sulphur is : 0.7 : 0.4 or 7 : 4 In the second sample Mass of iron dissolved = 2.8 G Mass of iron sulphide formed = 4.4 g Mass of sulphuric reacted `= (4.4 - 2.8) = 1.6 g` Ratio of mass of iron and sulphur is : 2.8 : 1.6 or 7 : 4 Since the ratio by mass of Fe and S in the TWO samples is the same, this VERIFIES the Law of constant composition. |
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| 51909. |
At 400 K 1.5 g of an unknown substance is dissolved in solvent and the solution is made to 1.5 L. Its osmotic pressure is found to be 0.3 bar. Calculate the molar mass of the unknown substance. |
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Answer» Solution :`DeltaT_(b) = K_(b) m = K_(b) XX W_(2) xx 1000//M_(2) xx W_(1)` `M _(2) = K_(b)xx W_(2) xx 1000//Delta T _(b) xx W_(1)` `= 7.5 xx 0.75 xx 1000//0.15 xx 200 = 187.5 g mol ^(-1)` |
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| 51910. |
0.759 g of silver salt of a dibasic acid was ignited to form 0.463 g of the silver residue. Calculate the molecular mass of the acid. |
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Answer» Solution :Step 1: EQUIVALENT MASS of silver salt of the acid. Equivalent mass of silver salt of the acid `= ("Mass of silver salt")/("Mass of silver residue") xx 108` `= (0.759)/(0.463) xx 108 = 177` Step II. Equivalent mass of acid Equivalent mass of acid `=` Equivalent mass of silver salt - Equivalent mass of Ag + Equivalent mass of H `= 177 - 108 + 1 = 70` Step III. Molecular mass of acid Molecular mass of acid = Equivalent mass `xx` basicity `= 70 xx 2 = 140` |
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| 51911. |
0.759 g of a silver salt of a dibasic organic acid on ignition left 0.463 g metallic silver. The equivalent mass of the acid is : |
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Answer» 70 |
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| 51912. |
0.75 mole of solid A_(4)and 2 moles of gaseous O_(2)are heated in a sealed vessel, completely using up the reactants and producing only one compound. It is found that when the temperature is reduced to the initial temperature, the contents of the vessel exhibit a pressure equal to half the original pressure. What conclusions can be drawn from these data about the product of the reaction? |
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Answer» SOLUTION :First method: As both `A_(4)` and `O_(2)` are consumed `("MOLES of" A_(4))/("moles of" O_(2)) =0.75/2.0` `=(("moles of A")//4)/(("moles of O")//2) = 0.75/2.0` `=("moles of A")/("moles of O") = 3/4` Thus, the empirical FORMULA of the product is `A_3O_4`. Further, as 2 moles of `O_(2)`give 1 mole of `A_3O_4` (for gases, pressure `prop` mole at constant temperature and VOLUME), `A_3O_4` is also the molecular formula of the product. Second Method :`underset("0.75 mole")(A_(4)s) + underset("2.0 mole")(O_(2) g) to underset("1.0 mole")(A_(x)O_(y)(g))` Applying POAC for A atoms, `4 xx` moles of `A_(4) = x xx `moles of `A_(x)O_(y)` `4 xx 0.75 = x xx 1, x=3` Applying POAC for O atoms. `2 xx` moles of `O_(2)= y xx` moles of `A_(x)O_(y)` `2 xx 2 = y xx 1, y =4` In the following chapters, we shall apply the principle of atom conservation (POAC) along with the said rules in tackling the various problems encountered in chemical practice. |
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| 51913. |
0.75 g platinichloride of monoacid base on ignition gave 0.245 g platinum. The molecular mass of the base is : |
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Answer» `75.0` W = weight of platinichloride w = weight of platinum, then `(W)/(2E+410)=(OMEGA)/(195)`, `(0.75)/(2E+410)=(0.245)/(195)RARR E = 93.5` |
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| 51914. |
0.72 gm of an oxide of a metal M on reduction with H_(2) gave 0.64 g of the metal . The atomic weight of the metal is 64 . The empirical formula of the compound is |
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Answer» MO |
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| 51915. |
0.70 g of a sample of Na_(2)CO_(3).xH_(2)O were dissolved in water and the volume was made to 100 ml. 20 ml of this solution required 19.8 ml of (N/10) HCl for complete neutralisation. The value of x is |
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Answer» 2 `20xxN=19.8xx(1)/(10)impliesN=(19.8)/(200)` `(W)/(VxxM.wt)xx2=(19.8)/(200)impliesM.wt=141.4impliesx=2` |
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| 51916. |
0.7 moles of potassium sulphate is allowed to react with 0.9 moles of barium chloride in aqueous solutions. The number of moles of the substance precipitated in the reaction is |
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Answer» 1.4 MOLES of potassium chloride `CaCO_(3) rarr CaO +CO_(2)` |
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| 51917. |
0.7 g of Na_(2)CO_(3).xH_(2)O was dissolved in water to make 100 mL solution, 20 mL of this solution required 19.8 mL of 0.1 N HCl for complete neutralisation. The value of x is : |
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Answer» 5 |
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| 51918. |
Combustion of 0.6gm of an organic compound gave 1.17gm of carbondioxide. 0.84 gm of water. Vapour density of the compound is equal to 22.4. The compound contains carbon, hydrogen and nitrogen. Calculate the molecular formula. |
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| 51919. |
0.66 g of a compound gave 112 ml of nitrogen at STP in the Dumas method. The percentage of Nitrogen in the compound is |
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Answer» 25 |
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| 51920. |
0.32 g of metal gave on treatment with an acid 112 mL of hydrogen at NTP. Calculate the equivalent weight of the metal. |
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Answer» Solution :Mass of the metal taken (`w_(1)) = 0.65` g Mass of 224 ML of hydrogen LIBERATED `w_(2) = 2/(22400) XX 224 = 0.02 g` HENCE, Equivalent weight of the metal `=w_(1)/w_(2) =0.65/0.02 = 32.5` |
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| 51921. |
0.66 g of H_(3)PO_(2) will require x ml of 0.1M NaOH for complete neutralization. X is |
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Answer» 100 ml `(0.66)/(66)XX1=(x xx0.1)/(1000)xx1impliesx 100mL` |
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| 51922. |
0.635 g of a metal gives on oxidation 0.795g g of its oxide. Calculate the equivalent mass of the metal. |
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Answer» Solution :Mass of the METAL oxide = 0.795g Mass of the metal= 0.635g Mass of oxygen= 0.795 - 0.635 = 0.16g 0.16 g of oxygen has COMBINED with 0.635 g of a metal `:.` 8g of oxygen will combine with ` (8 xx 0.635)/(0.16)=31.75g " equiv"^(-1)` |
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| 51923. |
0.63 grams of a dioic acid required 50mL of 0.2M NaOH for neutralisation. Calculate the molarmass of acid . |
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Answer» |
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| 51924. |
0.6% solution of urea and 1.8% solution containing a solute (A) are isotonic with each other. Calculate the molecular weight of the solute (A). |
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Answer» SOLUTION :`X=0.6% y=1.8%, M_(x)="UREA" (60 g "mol"^(-1)),M_(y)=?` `(x)/(100)xx(100)/(M_(x))=(y)/(100)xx(100)/(M_(y))=(x)/(M_(x))=(y)/(M_(y))` `M_(y)=(y xx M_(x))/(x)=(1.8xx60 g "mol"^(-1))/(0.6)` `=180 g "mol"^(-1)` |
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| 51925. |
0.6 moles of K_(2)Cr_(2)O_(7) can oxidise- |
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Answer» 3.6 mol of `FeSO_(4)` to `Fe_(2)(SO_(4))_(3)` |
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| 51926. |
0.6 mol K_(2)Cr_(2)O_(7), in acid medium can oxidise : |
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Answer» 3.6 mol `FeSO_(4) " to " Fe_(2)(SO_(4))_(3)` |
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| 51927. |
0.6 g of an organic compound was Kjeldhalised and NH_(3) evolved was absorbed into 50 ml of semi-normal solution of H_(2)SO_(4) . The residual acid solution was diluted with distilled water and the volume made up to 150 ml. 20 mL of this diluted solution required 35 mL of (N)/(20) NaOH solution for complete neutralisation. Calculate the % of N in the compound. |
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Answer» Solution :WEIGHT of Organic compound = 0.6 g Volume of SULPHURIC acid taken = 50 mL Strength of sulphuric acid taken=0.5 N 20 ml of diluted solution of unreacted sulphuric acid was neutralised by 35 ml of 0.05 N Sodium laydroxide `"Strength of the diluted sulphuric acid"=(35 xx 0.05)/(20) = 0.0875 N` Volume of the sulphuric acid remaining after reaction with ammonia= `V_(1)` mL Strength of `H_(2)SO_(4)` = 0.5 N Volume of the diluted `H_(2)SO_(4)` = 150 mL Strength of the diluted sulphuric acid = 0.0875 N `V_(1)= (150 xx 0.087)/(0.5) = 26.25 ml` Volume of `H_(2)SO_(4)` consumed by ammonia = 50 - 26.25 = 23.75 mL 23.75 mL of 0.5 N `H_(2)SO_(4)` - 23.75 mL of 0.5 N `NH_(3)` The amount of NITROGEN present in the 0.6 g of organic compound `=(14 g)/(1000 mL xx 1 N) xx 23.75 xx 0.5 N=0.166 g` `"Percentage of nitrogen"=(0.166)/(0.6) xx 100=27.66%` |
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| 51928. |
0.6 g of a metal gives on oxidation 1 g of its oxide. Calculate its equivalent mass. |
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Answer» Solution :Mass of METAL = 0.6 Mass of metal oxide = 1 g `:.` Mass of OXYGEN = 1 - 0.6 = 0.4 g 0.4 g of oxygen combines with 0.6 g of metal `:.` 8 g of oxygen will combine with `0.6/0.4 XX 8` Equivalent mass of the metal = 12 g `eq^(-1)` . |
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| 51929. |
0.5g sample of iron containg mineral mainly in the form of CuFeS_(2) was reduced suitably to vonvert all the ferruic ions into ferrous ions (Fe^(3+) rarr Fe^(2+)) and was obtained as solution. In the absenceof any interferringradical, the solution. In the absenceof any interfeerring radical, the solution required 42mL of interferring radical, the solution required 42mL of 0.1M K_(2)Cr_(2)O_(7) for tirtration. Calculate% of CuFeS_(2) in sample. |
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Answer» |
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| 51930. |
0.5g of an organic compound gave 62.2ml of N_(2) by Duma's method. Calculate the percentage of N in this compound. |
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Answer» `= (28 xx 62.2 xx 100)/(22400 xx 0.5) = 15.55%` |
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| 51931. |
0.5g of a metal on oxidation gave 0.79g of its oxide. The equivalent mass of the metal is |
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Answer» 10 |
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| 51932. |
0.59g of the silver salt of an organic acid (mol.wt. 210) on ignition gave 0.36 g of pure silver. The basicity of the acid is [AW of Ag = 108] |
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Answer» 1 BASICITY `= ("Mol. wt")/("EG. wt")=(210)/(70)=3` |
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| 51933. |
0.59g of the silver salt of an organic acid (mol. wt. 210) on ignition gave 0.36g of pure silver. The basicity of the acid is [AW of Ag = 108] |
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Answer» 1 |
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| 51934. |
0.59 g of an organic compound produces 112 mL nitrogen at NTP. The percentage of nitrogen in the compound through Duma's method is |
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Answer» 23.7 `%` of `=N=(28xx"VOLUME of " N_(2)" at STP")/(22400 xx "Mass of an ORGANIC compound")xx100` `=(28xx112xx100)/(22400xx0.59)=23.7`. |
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| 51935. |
0.548 g of the metal reacts with dilute acid and liberates 0.0198 g of hydrogen at S.T.P. Calculate the equivalent mass of the metal. |
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Answer» Solution :0.548 g of the METAL DISPLACES 0.0198 g of hydrogen The mass of the metal which will DISPLACE 1.008 g of hydrogen ` = (1.008 xx 0.548)/( 0.0198)g` of metal The equivalent mass of the metal `=27.90 g"equiv"^(-1)` . |
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| 51936. |
0.532 g of the chloroplatinate of a diacid base gave 0.195 g of the platinum residue on ignition. Calaculate the molecular mass of the base. |
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Answer» Solution :Mass of the base chloroplatinate taken `(W) = 0.532 G` Mass of the platinum residue `(x) = 0.195 g` step I. Equivalent mass of the base Equivalent mass of the base `= (1)/(2) [(W)/(x) xx 195 - 410] = (1)/(2) [(0.532)/(0.195) xx 195 - 410]` `= (1)/(2) [532 - 410] = 61` Step II. Molecular mass of the base Molecular mass of the base = Equivalent mass `xx` ACIDITY `= 61 xx 2 = 122` |
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| 51937. |
0.532 g of the chloroplatinate of an organic base (mol.wt 24 D) gave 0.195 g of Pt on ignition. Then the number of nitrogen atoms per molecule of the base is |
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Answer» 1 `(0.532)/(2E+410)=(0.195)/(195)RARR E=61` |
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| 51938. |
0.530 g of an organic compound gave 0.90 g of BaSO_(4) in carius determination of sulphur. Calculate the percentage of sulphur. |
| Answer» SOLUTION :`S = 23.32%` | |
| 51939. |
0.515 g of an organic compound containing phosphorus gave 0.214 g of magnesium pyrophosphate in Carius method for the estimation of phosphorus. Calculate the percentage of phosphorus in the given organic compound. |
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Answer» <P> Solution :MASS of organic compound `=0.515 g`Mass of MAGNESIUM pyrophosphate, `Mg_(2)P_(2)O_(7)=0.214 g` `{:(Mg_(2)P_(2)O_(7),equiv,2P),(22g,,62g):}` 222g of `Mg_(2)P_(2)O_(7)` are obtained from PHOSPHORUS `=62 g` 0.214 g of `Mg_(2)P_(2)O_(7)` are obtained from phosphorus `=(62xx0.214)/222=0.0597 g` Percentage of phosphorus `=("Mass of phosphorus")/("Mass of organic compound")xx100=(0.0597xx100)/0.515=11.6`. |
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| 51940. |
0.50 g of an organic compound was kjeldahlished. The ammonia evolved was passed in 50 cm^(3) of 1 N H_(2)SO_(4). The residual acid required 60 cm^(3) of N//2 NaOH solution. Calculate the pecentage of nitogen in the compound. |
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Answer» Solution :Step I. calculation of volume of UNUSED acid Volume of `NaOH` solution required `=60 cm^(3)` Normality of `NaOH` solution `=1//2 N` Normality of `H_(2)SO_(4)` solution `= 1N` Volume of unused acid can be calculated by applying normality equation `ubrace(N_(1)V_(1))_("Acid")=ubrace(N_(1)V_(1))_("Base") 1xxV=1/2xx60=30 cm^(3)` Step II. Calculation of volume of acid used Volume of acid added `=50 cm^(3)` Volume of unused acid `=30 cm^(3)` Volume of acid used `=(50-30)=20 cm^(3)` Step III. Calculating of percentage of nitrogen MASS of compound `=0.50 g` Volume of acid used `=20 cm^(3)` Normality of acid used `= 1 N` PERSENTAGE of `N=(1.4xx"Volume of acid used"xx"Normality of acid used")/("Mass of the compound")` `=(1.4xx20xx1)/0.50=56%` |
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| 51941. |
0.50 g of an organic compound was Kjeldahlished. The ammonia evolved was passed in 50 cm^(3) of 1N H_(2)SO_(4). The residual acid required 60 cm^(3) of N // 2 NaOH solution. Calculate the percentage of nitrogen in the compound. |
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Answer» Solution :Step 1. CALCULATION of volume of unused acid `"Volume of NaOH solution required" = 60 cm^(3)` `"Normality of NaOH solution" =(1)/(2)N` `"Normality of" H_(2)SO_(4) "solution"= (1)/(N)` Volume of unused acid can be calculated by applying normality equation `underset("Acid")(underbrace(N_(1)V_(1)))=underset("BASE")(underbrace(N_(1)V_(1)))` `1 xx V=(1)/(2) xx 60 = 30 cm^(3)` Step II. Calculation of volume of acid used `"Volume of acid added" = 50 cm^(3)` `"Volume of unused acid" = 30 cm^(3)` `"Volume of acid used"= (50-30) = 20 cm^(3)` Step III. Calculation of percentage of NITROGEN Mass of compound = 0.50 g `"Volume of acid used"=20 cm^(3)` Normality of acid used = 1 N `"Percentage of N" = (1.4 xx "Volume of acid used" xx "Normality of acid used")/("Mass of the compound")` `=(1.4 xx 20 xx 1)/(0.50)=50%` |
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| 51942. |
05 moles of NH_(4)HS(s) are taken in a container having air at 1 atm. On warming the closed container to 50^(@)C the pressure attained a constant value of 1.5 atm, with some NH_(4)HS(s) remaining unreacted. The K_(p) of reaction NH_(4)HS(s)hArrNH_(3)(g)+H_(2)S(g) "at" 50^@)C is: |
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Answer» `0.25` `P P` `2P=1.5-1` `2P=0.5` `P=0.25` `rArrK_(p)=0.25=0.0625` |
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| 51943. |
0.5 moles of gaseousnon-metallic X^(-) anions (having positive electron affinity) requires 806.4 kJ energy to get completely converted into gaseous X^(+) ions. Calculate Pauling's electronegativity of the elements X. Use Avogardo's no = 6 xx 10^(23) and 1eV = 1.6 xx 10^(-19)J. [Use the fact that, Pauling's electronegativity =("Mulliken's electronegativity")/(2.8) and Mulliken's electronegativity =("Ionisation energy+ Electron affinity")/(2)] |
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Answer» `:. IE` (energy absorbed) = aeV PER atom and `X +e^(-) rarr X^(-)` `:. EA` (energy RELEASED) =- beV per atom Now `(N)/(2)` of `X^(-)` will lose two elecctrons to give `(N)/(2)X^(+)` `:. X^(-) rarr X +E^(-)` `:.` energy absorbed `= + beV` per atom. `X rarr X^(+) +e^(-)` `:.` energy absorbed `=+ aeV` per atom. `:. a xx (N)/(2) + b xx (N)/(2) = (806.4xx 1000)/(1.6 xx 10^(-19)) eV` or `(a+b) = (806.4 xx 1000 xx 2)/(1.6 xx 10^(-19) xx 6 xx 10^(23)) = (806.4 xx 2)/(1.6 xx 60) = 16.8` `xhi_(m) = (IE +EA)/(2) = (16.8)/(2) = 8.4` `:.` Mulliken's electronegativity = 8.4 Pauling's electronegativity `= (8.4)/(2.8) = 3.0` |
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| 51944. |
0.5 mole of S_8 and 0.5 mol of P_4 have same number of polyatomic molecules |
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Answer» |
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| 51945. |
0.5 mole of P_(4)IO_(10) contains |
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Answer» 80 gm oxygen 1 mole `P_(4)O_(10)=284g` `284 g P_(4)O_(10)` contains 124 g phosphorous `284g P_(4)O_(10)` contain 160 g oxygen 0.5 moles `P_(4)O_(10)-62 g P - 80 g` oxygen (2 gram atom) (5 gram atom) |
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| 51946. |
0.5 mole of ethanol is mixed with 1.5 mole of water. Then the mole fraction of ethanol and water are …………….. |
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Answer» SOLUTION :`0.25, 0.75` MOLE fraction of ethnol `= ("NUMBER of moles of ETHANOL")/("Total number of moles of ethanol and WATER")` `=(0.5)/( 1.5 + 0.5) = (0.5)/(2.0 = 0.25` Mole fraction of water `= (1.5)/( 2.0) = 0.75` |
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| 51947. |
0.5 mole of each of H_(2),SO_(2) and CH_(4) are kept in acontainer. A hole was made in the container. After 3 hours, the order of partial pressures in the container will be |
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Answer» `p_(SO_(2)) gt p_(CH_(4)) gt p_(H_(2))` `prop sqrt((1)/("Mol mass"))`,therefore, RATES of diffusion will he in the ORDER : `H_(2) gt CH_(4) gt SO_(2)`. Hence, amounts left will be in the order `H_(2) lt CH_(4) lt SO_(2)` or `SO_(2) gt CH_(4) gt H_(2)`. Their partial pressures will also be in the same order. |
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| 51948. |
0.5 mole of ethanol is mixed with 1.5 mole of water. Then the mole fraction of ethanol and water are .... |
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Answer» `0.75 ,0.25` SOL :molefractionof ethanol `=( "Numberof MOLESOF ethanol ")/(" TOTALNUMBEROF molesof ethanoland WATER ")` `= ( 0.5 )/( 1.5+0.5 ) = ( 0.5) /( 2.0 ) = 0.25` Molefractionof water`=( 1.5 )/( 2.0 ) =0.75` |
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| 51949. |
0.5 mole of calcium carbonate id decomposed by an aqueous solution containing 25 % JCl by mass. Calculate the mass of the solution consumed. |
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Answer» `underset("1 mol")(CaCO_(3))+underset("2 mol")(2HCl)rarrCaCl_(2)+H_(2)O+CO_(2)` 1 mole of `CaCO_(3)` react with HCl = 2 mol 0.5 mole of `CaCO_(3)` react with HCl = 1 mol = 36.5 g Mass of HCl solution REQUIRED ot consumed `= 36.5 xx (100)/(25) = 146 g`. |
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| 51950. |
0.5 mole each of two ideal gases A (C_(v.m)= (5)/(2)R) and B (C_(v.m)=3R) are taken in a container and expanded reversibly and adiabtically, during this process temperature of gaseous mixture decreases from 350K and 250K. Find DH (in cal/mol) for the process |
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Answer» `-100R` `DELTA H = nCp Delta T = 1 xx 3.75 xx (-100) = - 375R` |
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