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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.
| 51651. |
100ml of water at 20^@C and 100ml of water at 40^@C are mixed in calorimeter until constant temperature reached. Now temperature of the mixture is 28^@C. Water equivalent of calorimeter is |
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Answer» 50J |
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| 51652. |
100ml of O_2 diffuses in 100 minutes, under similar conditions one litre of H_2 diffuses in |
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Answer» 100min |
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| 51653. |
100mL of H_(2)O_(2) solution having volume strength 11.35 V is mixed with 500mL of 0.5M KI solution to liberate I_(2) gas such that equilibria gets established. All the I_(2) gas liberated is dissolved to form 500mL solution. 200ml of the solution required 50mL of 2/(3)M hypo solution. Calculate volume strength fo remaining H_(2)O_(2) mixture. Round off your answer. |
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| 51654. |
100mL of each hydrogen and oxygen was mixed and fired. What will be the final volume of the mixture when measured at the original conditions? |
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Answer» Solution :The storichiometric equation is given as `2H_(2)(g)+O_(2)(g)to2H_(2)O(1)` 2vol1vol 0 vol Gay-Lussac.s coefficients 100ML 100mL 0mLat START 0mL 50ML 0mL after the reaction The composition of the final mixture is that it has only unreacted OXYGEN volume =50ml |
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| 51655. |
100mL of each 0.2M solutions of H_(2)SO_(4) and HCl are mixed. Calculate the normality of the mixture. |
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Answer» Solution :Milliequivalents of `H_(2)SO_(4)=100xx0.2xx2=40` Milliequivalents of HCL `=100xx0.2xx1=20` TOTAL number of milliequivalents of acid in the MIXTURE =40+2=60 Total volume of mixture =100+100=200ml Normality of the mixture `=("M EQ")/("Volume")=(60)/(200)=0.3 eq L^(-1)` |
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| 51656. |
100mL of a gaseou mixture has 20mL ethylence and the remaining oxygen. The mixture Is fired and cooled to room temperature. Determine the final composition of gaseou mixture. |
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| 51657. |
1.00litre sample of a mixture of CH_(4(g)) and O_(2(g)) measured at 25^(@)C and 740 t or r was allowed to react at constant pressure in a calorimeter which together with its contents had a heat capacity of 1260cal//K. The complete combustion of the methane toCO_(2) and H_(2)O caused a temperature rise in the calorimeter of 0.667K. What was the mole per cent of CH_(4) in the original mixture ? (DeltaH_(comb.)^(@)(CH_(4))=-215kcal mol^(-1)) |
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| 51658. |
100kg of a water sample contains 6g of magnesium sulphate, 10.2g of calcium sulphate and 16.2 g of calcium bicarbonate only. Calculate permanent hardness, temporary hardness and total hardness. |
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Answer» Solution :`100 Kg (10^5g)`of given water SAMPLE contains 6 g of `MgSO_4, 10.2 g CaSO_4` and `16.2g CA(HCO_3)_2` . Weight of `MgSO_4`in `10^6`g water `= (6)/(10^5) xx 10^6= 60g` 120 g of `MgSO_4` = 100 g of `CaCO_3` 60 g of `MgSO_4 =60/120 xx 100 = 50 g` of `CaCO_3` Degree of HARDNESS due to `MgSO_4` = 50ppm weight of `CaSO_4` in `10^6g` of water = `10.2 xx 10^6//10^5 = 102g` 136 g of `CaSO_4` = 100 g of `CaCO_3` 102 g of `CaSO_4 = 102 xx 100//136 = 75` ppm Degree of hardness due to `CaSO_4 = 75` ppm Weight of `Ca(HCO_3)_2` present in `10^6` gm of water ` = 16.2 xx 10^6//10^5 = 162 g ` 162 g of `Ca(HCO_3)_2 -= 100 g `of `CaCO_3` Degree of hardness due to `Ca(HCO_3)_2 = 100` ppm Temporary hardness of water sample = 100ppm Permanent hardness of water sample = 50 + 75 = 125 ppm TOTAL hardness of water sample = 100+125=225 ppm |
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| 51659. |
100kg of a sample of water contained 6 of MgSO_(4) The degree of |
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Answer» 6 ppm HARDNESS of `MgSO_(4)=(6g)/(10^(5)g)xx10^(6)=60ppm` hardness in term of `CaCO_(3)=60xx(100)/(120)=50` ppm |
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| 51660. |
100Kg of a sample of water contained 6 g of MgSO_4 . The degree of hardness |
| Answer» ANSWER :D | |
| 51661. |
100grams of a hydrocarbon has 93.71 g of carbon. What is its empirical formula of the hydrocarbon? |
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| 51662. |
100cm^3of a given sample of H_2O_2gives 1000cm^3ofO_2 at S.T.P. The given sample is |
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Answer» `10% H_2O_2` |
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| 51663. |
100cc of CO_2 gas is diffused in 25 seconds through a porous membrane. How much time does the same volume fo sulphur dioxide take to diffuse ? |
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| 51664. |
1000L of air a STP was dissolved in water and required 2.5xx10^(-5) moles of KMnO_(4) for complete reaction of SO_(2) as pollutants. Thus , SO_(2) content in air is |
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Answer» `1.4ppm` `underset(+7)(2MnO_(4)^(-))+underset(+4)(2H_(2)SO_(3)) rarr underset(+6)(5H_(2)SO_(4))+underset(+2)(2Mn^(2+))` `2 mol ES MnO_(4)^(-) = 5 mol e H_(2)SO_(3) or SO_(2)` `2.5xx10^(-5)` mole `MnO_(4)^(-) = 5/2 XX 2.5xx10^(-5) mol es SO_(2)` `=6.25xx10^(-5)mol es SO_(2)` At STP, `6.25xx10^(-5) mol es = 6.25xx10^(-5)xx22.4L SO_(2)` `=1.4xx10^(-3)L` In `1000L` air, `SO_(2)=1.4xx10^(-3)L` In `10^(6)L(ppm), SO_(2) = (1.4xx10^(-3)xx10^(6))/(10^3) = 1.4ppm`. |
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| 51665. |
1000g of a water sample contained a calcium salt equivalent to 1g of CaCO_3. The hardness of that water sample is |
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Answer» 100 ppm |
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| 51666. |
1000g of a water sample contained a calcium salt equivalent to 18 of CaSO_(4) The hardness of that water sample is |
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Answer» 100 ppm |
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| 51667. |
100ml of KMnO_4 salution is exactly reduced by 100ml of 0.5 M oxalic acid under acidic condition. The molarity of KMnO_4 solution is |
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Answer» 0.1M |
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| 51668. |
1000 mL O_(2) at NTP was passed through Siemen's ozonizer so that the volume is reduced to 888 mL at same condition. Ozonized oxygen is passed through KI solution. Liberated I_(2) was titrated with 0.05 N hypo. Calculate volume of hypo used. |
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| 51669. |
10.00 mol of pure, 1-butyne was taken in a flask with some basic catalyst, sealed and heated to 500K where it isomerised to produce 2-butyne and 1,3-butadiene. At equlibrium 3.0 mole of 2-butyne and 5.5 mole of 1.3-butadiene were found to be present in the flask. If 2.0 mole of 1.3-butadiene wer added further to the above equilibrium mixture. What would be the new equilibrium composition when equilibrium was established again? 1-butyne 2-butyne 1,3-butadiene |
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Answer» 1.80moles 3.60moles 3.60moles |
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| 51670. |
100 ml solution having 0.2 M HA (weak acid, K_(a), = 1.0x10^(-5) ) and 0.2 N NaA, 200 ml of 0.1 M NaOH has been added. Furthermore, diluted to IL. Which of the following statement is correct? |
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Answer» Initially , the SOLUTION has pH equal to 5, that is before additions of NaOH ` pH =5+ LOG ""(0.2)/( 0.2)= 5` (b)after dilution to1 LT, ` [HA] =0.02 N, [NaA ]= 0.02 N` ` [NaOH ]=0.02 N` ` {:( NaOH +, HA to , NaA + H_2O) , (0.02 N , 0.02N , -),( -,-,0.02N):}` ` therefore` Due to addition of NaOH , buffer disappeared & salt is formed ` pH =7+ ( 5)/(2) +(1)/(2) log (4 XX 10 ^(-2)) = 8.8 ` |
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| 51671. |
100 mL solutionof FeC_(2)O_(4) and FeSO_(4) is compleltely oxidized by 60 mL of 0.2M KMnO_(4) in acid medium. The resulting soltuion is thenreduced by Zn and dil. The reduced soltuion is again oxidized completely by 40 mLof0.2M KMnO_(4). Calculate normality of FeC_(2)O_(4) and FeSO_(4) in mixture. |
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| 51672. |
100 ml solution contains 12 mg MgSO_(4). The concentration of solution is |
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Answer» `10^(-3)M` ppm `(w_(1))/(w_(2))xx10^(6)=(12xx10^(-3))/(100)xx10^(6)=120`ppm For dilute solution, m = M |
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| 51673. |
100 mL sample of hard water is passed through a column of the ion exchange resin RH_(2). The water coming off the column requires 15.17 mL of 0.0265 M NaOH for its titration. What is the hardness of water as "ppm" of Ca^(2+)? |
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| 51674. |
100 mL solution consists 4 g caustic soda. The normality of the solution is : |
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Answer» `1.0` |
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| 51675. |
100 mL of tap water containing Ca(HCO_(3))_(2) was titrated with N/50 HCl with MeOH as indicator . If 30 mL of HCl were required , calculate the temporary hardness as parts of CaCO_(3) per 10^(6) parts of water . If your answer is 'a x 100' , what is the value of 'a'. |
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| 51676. |
100 ml of sample of water requires 3.92 mg of K_2 Cr_2O_3 in presence of H_2SO_4 for the oxidation of dissolved organic matter present in it. The COD of the water sample in ppm |
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Answer» 3.1 |
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| 51677. |
100 ml of sample is removed from an aqueous solution saturated with CaSO_(4) at 25^(0)C. The water is completely evaported from the sample and deposit of 0.24 g of CaSO_(4) is obtained. The Ksp of CaSO_(4) at 25^(0)C will be. |
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Answer» `3.115 xx 10^(-4)` According to the given DATA, solubility of `CaSO_(4)` is `0.24 g` PER `100 ml` `[CA^(+2)] = [SO_(4)^(2-)] = 0.01765 M` `Ksp = [Ca^(+2)] [SO_(4)^(2-)]` |
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| 51678. |
100 mL of phosphorus pentachloride is totally decomposed to its trichloride at 1 atm and 546^(@)C. How many molecules of chlorine are formed? |
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Answer» Solution :DECOMPOSITION of phosphorus pentachloride is given as `PCl_(5)rarrPCl_(3)+Cl_(2)` `"1 mole of "PCl_(5)="1 mole of"Cl_(2)` `"1 vol of "PCl_(5)="1 vol FO Cl"_(2)` `"100 mL of "PCl_(5)="100 mL of "Cl_(2)` `{:("Given conditions","STP conditions"),(P_(1)="1 atm",P_(2)="1 atm"),(T_(1)="819 K",T_(2)="273 K"),(V_(1)="100 mL",V_(2)=?):}` The VOLUE of `Cl_(2)` MEASURED at STP `V_(2)=(P_(1)V_(1))/(T_(1))xx(T_(2))/(P_(2))=(1xx100xx273)/(819xx1)="33.33mL"` `"22400 mL of "Cl_(2)" at STP"=` `6.022xx10^(23)" molecules"` `"33.33 mL of "Cl_(2)" at STP"=?` Number of molecules of chlorine obtained `=(33.33)/(22400)xx6.022xx10^(23)=8.96xx10^(20)` |
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| 51679. |
Phosphine on decomposition produces phosphorus and hydrogen. When 100 ml of phosphine are decomposed the change in volume under laboratory conditions is |
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Answer» 50ML increase |
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| 51680. |
100mL of om at STP was passed through 100 ml of 10 volume H_(2)O_(2)solution. What is the volume strength ofH_(2)O_(2) after the reaction? |
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Answer» 9.5 initial `M_(H_2O_2) = 10/11.2 = 10/11.2 M` NUMBER of MOLES of `O_2 = 0.1/22.4` number of moles of `O_2 = 10/11.2 xx 0.1` ACCORDING to stoichiometric equation on mole of `H_2O_2` produce 1 mole of `O_3` number of moles of `H_2O_2` consumed `=1/224," remains" = 19/224` onemolarity = 11.2 (W/V)% `19/224 xx 0.1 --------- ?` |
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| 51681. |
100 mL of HCI gas at 25^(@)C and 740 mm pressure were dissolveed in one litre of water. Caculate the pH of a solution. Given, V.P. of H_(2)O at 25^(@)C is 23.7 mm. |
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| 51682. |
100 mL of H_(2)O_(2) is oxidized by 100 mL of 1M KMnO_(4) in acidic medium (MnO_(4)^(-) reduced to Mn^(+2) ) 100 mL of same H_(2)O_(2) is oxidized by v mL of 1M KMnO_(4) in basic medium (MnO_(4)^(-)reduced to MnO_(2)). Findthe value of v: |
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Answer» Solution :In acidic medium Eqts of `H_(2)O_(2)` = Eqts of `KMnO_(4)` `(100)/(1000)xxMxx2=(100)/(1000)xx1xx5impliesM=2.5` In basic medium Eq. of `H_(2)O_(2)` = Eqts of `KMnO_(4)` `(100)/(1000)xx2.5xx2=(V)/(1000)xx1xx3impliesV=(500)/(3)` |
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| 51683. |
100 mL of each pH = 3 and pH = 5 solutions of a strong acid are mixed. What is the resultant pH ? |
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Answer» Solution : PH of first solution `= 3, [H^(+) ]=10^(-3)=N_1` pH of second solution `= 5, [H^(+) ]=10^(-5) = N_2` Final PROTON concentration is GIVEN by the equation:`[H^+]=(V_1 N_1+V_2 N_2)/(V_1 +V_2)` Substituting the values` [H^(+)] =(100 XX 10^(-3) xx 10^(-5))/( 100+100 ) = 5.05 xx 10^(-4)` M pH of the resultant mixture `= -log (5.05 xx 10^(-4) ) = 4 - log 5.05 = 3.3` |
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| 51684. |
100 mL of ethano (d = 0.78g cc^(-1)) I made upto a litre with pure water. Caculate the molality. |
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| 51685. |
100 mL of each acetylene and oxygen are mixed. The mixture is strongly heated to complete the reaction and colled back to room temperature. What is the maximum volume of carbondioxide obtained. Why? |
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Answer» Solution :The stoichiometric equation for burning of acetylene is given as `2C_(2)H_(2)+5O_(2)rarr4CO_(2)+2H_(2)O` `"2 moles of "C_(2)H_(2)="5 moles of "O_(2)` `"2 volumes of "C_(2)H_(2)=5" volumes of "O_(2)` `"100 mL of "C_(2)H_(2)=250ML" of "O_(2)` `"40 mL of "C_(2)H_(2)=100 mL " of "O_(2)` HENCE the limiting REAGENT is oxygen `"5 moles of "O_(2)="4 moles of "CO_(2)` `"5 volumes of "O_(2)="4 volumes of "CO_(2)` `"100 mL of "O_(2)=?` The maximum VOLUME of `CO_(2)` formed `=100xx(4)/(5)="80 mL"` This is because no SUFFICIENT `O_(2)` is available. |
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| 51686. |
100 ml of a sample of water requires 1.96 mg of potassium dichromate in the presence of 50% H_2SO_4 for the oxidation of dissolved organic matter in it. Calculate the chemical oxygen demand. |
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Answer» Solution :One gram equivalent of any OXIDANT can give, one equivalent weight of OXYGEN (8 G) `49 g K_2Cr_2O_7 to 8g " Oxygen"` ` 1.96 xx 10^(-3) g K_2Cr_2O_7 to (8 xx 1.96 xx 10^(-3))/(49)=0.32 mg ` Chemical oxygen demand of the given 100 ml SAMPLE of water `=(0.32)/(0.1) = 3.2 ppm` |
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| 51687. |
100 ml of a sample of water requires 1.96 mg of potassium dichromate in the presence of 50% H_(2) SO_(4) for the oxidation of dissolved organic matter in it. The chemical oxygen demand is 8x xx 10^(-1) ppm. The value of x is _____. |
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| 51688. |
100 ml of a sample of water requires 0.98mg of K_(2)(M.W.= 294 )in presence of the H_(2)SO_(4) for the oxidation of dissolved organic matter in it. The COD of the water sample is |
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Answer» 78.4 PPM |
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| 51689. |
100 mL of a gas is stored over mercury in mercury manometer at 27^(@)C radius of inside column is r and that of outside is R and initially mercury level is equal in both column (a) If R=r. Find the new temperature (in K) if due to change in temperature outside level of mercury is raised by 20 mm (i) Assume volume of gas remain constant by some experimental means rArr T_(1) (ii) Volume does not remain constant (pi r^(2) = 10 cm^(2)) rArr T_(2) (b) If R = 2r and inside level falls by 80 mm rArr T_(3) Give the answers by adding T_(1),T_(2),T_(3) to the nearest integer. |
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| 51690. |
100 ml of 2M HCl solution completely neutralises 10 g, of a metal carbonate. Then the equivalent weight of the metal is |
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Answer» 50 |
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| 51691. |
100 mL of 0.3 N HCl were mixed with 200 mL of 0.6 N H_(2)SO_(4) solution. The final normality of acid was : |
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Answer» 0.4 N |
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| 51692. |
100 mL of 0.2 N HCl solution is added to 100 mL of 0.2 N AgNO_(3) solution. The molarity of nitrate ions in the resulting mixture will be : |
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Answer» 0.05 M |
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| 51693. |
100 ml of 0.1M aqueous ferric alum solution is completely reduced with Iron to give Ferrous Ions. The volume of 0.1M acidified KMnO_4solution required for complete reaction with ferrous Ions in the solution will be |
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Answer» SOLUTION :`Fe^(+3)+Ferarr2Fe^(+2)` `100xx0.1 xx2=0.5xxV` `V=(100xx0.2)/(0.5)=200/5=40` ml |
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| 51694. |
100 ml of 0.1 M HCl and 100 ml of 0.1 M HOCN are mixed . What is the concentrationof .. OCN ^(-)..,and pH of the solution ?(K_a=1.2 xx 10^(-6) ) |
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Answer» ` 1.2 xx 10^(-6), 1.3 ` `[HOCN]=( 100xx 0.1 )/( 200) = 0.05` ` K_a =([H^(+) ][OCN^(-) ])/( [HOCN])=(1.2 xx 10 ^(-6) xx 0.05 )/( 0.05 ) ` ` = [OCN^(-) ]=1.2 xx 10^(-6) ` |
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| 51695. |
100 ml of 0.1 M HCl and 100 ml of 0.1 M HOCN are mixed then(K_a= 1.2 xx 10^(-6)) |
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Answer» `OCN^(-) `concentration in the solution is `1.2 xx10^(-6) ` ` K_a RARR ([H^(+)][OCN^(-)])/( [HOCN]) ` ` 1.2 xx 10 ^(-6)= (5xx 10 ^(_2)[OCN^(-) ])/( 5 xx10 ^(-2)) ` `[ OCN^(-)]= 1.2 xx 10 ^(-6) ` Approximation method, WEAK acid conribution is neglected. ` [H^(+) ] =(100 xx 0.1 +0)/( 200) = 0.05 therefore pH =1.3 ` |
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| 51696. |
100 mL of 0.01 M XO_(4)^(-) is reduced to X^(n+) by 100 mL of 0.05 M Fe^(2+) in acidic medium. Thusoxidation state of X in X^(n+) is ______________ |
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| 51697. |
100 ml of 0.01 M KMnO_(4) oxidizes 10 ml of H_(2)O_(2) sample in acidic medium. The volume strength of H_(2)O_(2) sample is |
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Answer» 11.2 Vol `100xx0.01xx5=10xxNimpliesN=0.5` IMPLIES V.S. = `Nxx5.6=2.8 vol` |
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| 51699. |
100 mL each of 1NH_(2)O_(2) " and " 11.2 V H_(2)O_(2) solution are mixed, then the resultant solution will be : |
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Answer» `3M H_(2)O_(2)` `N_(1)V_(1)+N_(2)V_(2)=N_(R )(V_(1)+V_(2))` `1xx100+2xx100=N_(R )xx200` `N_(R )=(3)/(2)=1.5` Strength `=N xx " Equivalent MASS"` `=1.5xx17=25.5 g//L` |
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| 51700. |
100 mL , each of 0.25 M NaF and 0.01 5 Ba (NO)_3 are mixed K_(sp)of BaF_2= 1.7 xx 10^(-6) |
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Answer» A ppt is formed ` therefore PPt `is formed |
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