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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.
| 901. |
How many moles of electrons weigh one kilogram?A. `6.023 xx 10^(23)`B. `(1)/(9.108)xx10^(31)`C. `(6.023)/(9.108)xx10^(54)`D. `(1)/(9.108 xx 6.023)xx10^(8)` |
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Answer» Correct Answer - D 1 mole of electron weighs `= 9.108 xx 10^(-31) xx 6.022 xx 10^(23)kg` No. of moles of electrons in 1 kg `= (1)/(9.108xx10^(-31)xx6.022xx10^(23))` `=(1)/(9.108xx6.022)xx10^(8)`. |
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| 902. |
How many moles of ptoton weigh 1 Kg?A. `10^(3)`B. `N_(A)`C. `(1)/(N_(A)) xx 10^(3)`D. `(10^(8))/(9.11 xx 6.022)` |
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Answer» Correct Answer - A |
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| 903. |
One atomic mass unit in kilogram is:A. `(1)/(N_(A))`B. `(12)/(N_(A)`C. `(1)/(1000N_(A))`D. `(1000)/(N_(A)` |
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Answer» Correct Answer - C |
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| 904. |
How many moles of electrons weigh one kilogram?A. `6.023 xx 10^(23)`B. `(1)/(9.108) xx 10^(31)`C. `(6.023)/(9.108) xx 10^(54)`D. `(1)/(9.108 xx 6.023) xx 10^(8)` |
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Answer» Correct Answer - D |
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| 905. |
What weight of `AgCl` will be precipitated when a solution containing `4.77 g NaCl` is added to a solution of `5.77 g` of `AgNO_(3)`. |
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Answer» Correct Answer - 4.879 gm `L.R` `darr` `AgNO_(3)+NaCl to NaNO_(3)+AgNO_(3)` `5.77 gm " " 4.77 gm ` ` 0.034 "mole" " " 0.082 "mole"` wt. of `AgCl=0.034xx143.5=4.879 gm` |
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| 906. |
21.6 g of silver coin is dissolved in `HNO_(3)`. When NaCl is added to this solution, all silver is precipitated as AgCl. The weight of AgCl is found to be 14.35g then % silver in coin is: `Ag + HNO_(3) overset(NaCl)to AgCl`A. 0.5B. 0.75C. 1D. 0.15 |
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Answer» Correct Answer - A |
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| 907. |
Haemoglobin contains 0.25% iron by weight , the molecular weight of Haemoglobin is 89600, calculate the weight (in g) of `K_(4)[Fe(CN)_(6)]` that can be produced if all the iron atoms from 4.48 kg haemoglobin are converted into `K_(4)[Fe(CN)_(6)]` through a series of reaction . Give your answer to the nearest integer. |
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Answer» Correct Answer - 74 |
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| 908. |
Concentrated `HNO_(3)` is `63%` `HNO_(3)` by mass and has a density of `1.4g//mL`. How many millilitres of this solution are required to prepare `250mL` of a `1.20 M HNO_(3)` solution ? |
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Answer» Correct Answer - 200 |
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| 909. |
Calculate the volume occupied by 200 g of `SO_(3)` gas at STP and the number of molecules present in it. |
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Answer» 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 is 22.4 L, at STP Volume occupied by the given amount of gas = `2.5 xx 22.4 = 56` L Number of molecules present in 1 mole of gas = ` 6.023 xx 10^(23)` Number of molecules present in the given amount of gas = `2.5 xx 6.023 xx 10^(23) = 15.05 xx 10^(23)` molecules |
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| 910. |
When a fully blown balloon is subjected to sudden bursting. What do you observe? Justify your observation. |
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Answer» (i) sudden expansion of the gas present inside the balloon (ii) change in energy of the system due to the sudden expansion of the gas (iii) effect of the change in energy on temperature (iv) effect of this change in temperature on the surrounding (v) phase transition of a vapour present in the surrounding |
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| 911. |
If 15 mg of `N_(2)O_(3)` is added to `4.82 xx 10^(20)` molecules of `N_(2)O_(3)`, the total volume occupied by the gas at STP isA. `0.044` LB. ` 0.022` LC. ` 0.22` LD. ` 0.44` L |
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Answer» Gram molecular weight of `N_(2)O_(3)` is 76 g and 76 g of `N_(2)O_(3)` contains `6.023 xx 10^(23)` molecules ` = (76 xx 4.82 xx 10^(20))/(6.023 xx 10^(23)) g = 0.061` g 15 mg of ` N_(2)O_(3) = 0.015` g `:." Total weight of "N_(2)O_(3) = 0.061 + 0.015 = 0.076` g 76 g occupies 22.4 L at STP. 0.076 g occupies ? L at STP. ` = (0.076 xx 22.4)/(76) = 0.022` L |
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| 912. |
A solution is prepared by dissolving `9.8" g of "H_(2)SO_(4)` in 54 g of water. What is the mole fraction of `H_(2)SO_(4)`? |
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Answer» Weight of `H_(2)SO_(4) = 9.8` g GMW of `H_(2)SO_(4) = 98` Number of moles of ` H_(2)SO_(4)` present in the solution = 9.8/98 = 0.1 Number of moles of `H_(2)O=54//18= 3` Mole fraction of `H_(2)SO_(4)= 0.1//(0.1+3)=0.032`. |
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| 913. |
X g of `CO_(2)` is prepared by the reaction of `CaCO_(3)` and HCl. How many grams of oxygen will be liberated, if X grams of `CO_(2)` completely take part in the process of photosynthesis ? 144 g of water is produced in the first reaction. Equation for photosynthesis is `CO_(2) + H_(2)O to C_(6)H_(12)O_(6)+O_(2)` |
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Answer» `CaCO_(3) + 2HCl to CaCl_(2) + CO_(2) + H_(2)O` `6CO_(2) + 6H_(2)O to C_(6)H_(12)O_(6) + 6O_(2)` Number of moles of water formed in the first reaction = `(144)/18 = 8` Number of moles of `CO_(2)` produced = 8 Number of moles of oxygen produced by using 8 moles `CO_(2)` is 8 . Weight of oxygen liberated ` = 32 xx 8 = 256` g |
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| 914. |
When one mole each of CO and `O_(2)` are made to react at STP, the total number of moles at the end of the reactions isA. `1.5` molesB. 1 moleC. 4 moleD. 2 moles |
| Answer» Correct Answer - A | |
| 915. |
Specific grabvity of 84% (w/w) pure `HNO_(3)` is 1.54. What volume of `HNO_(3)` is required to prepare one litre of 0.5 M `HNO_(3)` solutions? |
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Answer» Specific gravity of `HNO_(3) = 1.54` `M=w/"GMM"xx1000/V rArr0.5 = x/63 xx1000/1000 rArr x = 31.5`g `31.5" g of "NHO_(3)` is present in 1 L of solution. 84 g is present in 100 g of solution 31.5 g is present in ? g of solution ` = (31.5 xx 100)/84 = 37.5` g `"Density"="weight"//"volume"rArr1.54 =37.5//"volume" rArr V = 37.5//1.54 = 24.35` m L `V_(1)M_(1) = V_(2)M_(2)" "24.35`m L of given` NHO_(3)` is required. |
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| 916. |
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» `{:(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 produced from 11.2 L CO gas. |
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| 917. |
20 cc of a hydrocarbon, on complete combustion, gave 80 cc of `CO_(2) and " cc of " H_(2)O` at STP. The empirical formula of that compound isA. `C_(2)H_(5)`B. `C_(2)H_(6)`C. `C_(3)H_(8)`D. `C_(4)H_(10)` |
| Answer» Correct Answer - A | |
| 918. |
Empirical formula of a compound is `A_(2)B_(4)`. If its empirical formula weight is half of its vapour density, determine the molecular formula of the compound.A. `A_(4)B_(8)`B. `A_(8)B_(16)`C. `A_(2)B_(4)`D. `A_(3)B_(6)` |
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Answer» Let the vapour density of compound be x Empirical formula weight = x/2 Molecular formula weight = 2x `n=("Molecular formula weight")/("Empirical formula weight ")=(2x)/(x/2) = 4` Molecular formula = `4 xx A_(2)B_(4) = A_(8)B_(16)` |
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| 919. |
Calculate the weight of 80% pure limestone required to produce 11 g of `CO_(2)` gas. |
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Answer» `{:(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 the limestone is 80% pure, the weight of impure limestone required = `100/80 xx 25" g " = 31.2` g. |
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| 920. |
The empirical formula of a compound is `CH_(2)O`. If its vapour density is 90, find out the molecular formula of the compound. |
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Answer» Empirical formula=`CH_(2)O` Empirical formula weight = `= 12 + (1xx2) + 16 = 30` Vapour density (VD) = 90 Molecular wieght =`VD xx 2 = 180` Molecular formula = `("empirical formula")_(n)` `n=("Molecular weight")/("Empirical formula weight")=180/30 = 6` Thus, molecular formula = `(CH_(2)O)_(6) = C_(6)H_(12)O_(6)` In stoichiometry, we find out the quantitative relationship between the reactants and the products on the basis of balanced chemical equation. |
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| 921. |
In a `2.5` L flask at `27^(@)C` temperature, the pressure of a gas was found to be 8 atm. If ` 41 xx 10^(23)` molecules of the same gas are introduced into the container, the temperature changed to `T_(2)`. The pressure of gas is found to be 10 atm. Find out the value of `T_(2)`.A. 253 KB. 347 KC. 230 KD. 370 K |
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Answer» `6.023 xx 10^(23)` molecules correspond to 1 mole ` 2.41 xx 10^(23) ` molecules correspond to ? ` = (2.41 xx 10^(23) ) / (6.023 xx 10^(23)) = 0.4` moles Case (i) `._(1)V_(1) = n_(1).R_(1)." "T_(1)` ` 8.(2.5) = n_(1)(0.08).(300)` `n_(1) = (8(2.5))/((0.08)(300))` ` n_(1) = 0.833` moles Total moles, `n_(2) = 0.833 + 0.4` ` n_(2) = 1.233` Case (ii) ` P_(2)V_(2) = n_(2)R T_(2)` ` 10(2.5) = 1.233 (0.08). T_(2)` `T_(2) = 253.4` K |
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| 922. |
A ges at a pressure of 2.0 atm is heated from 0 to `273^(@) C` and the volume compressed to 1/4th of its original volume. Find the final pressure. |
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Answer» `{:("Initial condition",,"Final condition"),(V_(1)=V,,V_(2) = ¼ V),(T_(1) = 0^(@)C or 273 K,,T_(2) = 273^(@)C (or) 546 K),(P_(1) = 2" atm",,P_(2) =?):}` According to gas equation, `(P_(1)V_(1))/T_(1) = (P_(2)V_(2))/T_(2)` `P_(2) = (P_(1)V_(1))/T_(1) xxT_(2)/P_(2) =(2xxV)/273 xx546/V xx4,P_(2) = 16` atm |
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| 923. |
If a gas occupies 30 L at `27^(@)C` and 1 atm, what volume would it occupy at `227^(@)C` and 5 atm? |
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Answer» `{:("Initial condition",,"Final condition"),(V_(1) = 301,,V_(2) =?),(P_(1)=1" atm",,P_(2) = 5" atm"),(T_(1) = 27^(@)C = 300 K,,T_(2) = 227^(@)C = 500K):}` According to gas equation, `(P_(1)V_(1))/T_(1)= (P_(2)V_(2))/T_(2) V_(2)=(P_(1)V_1)/T_(1) xxT_(2)/P_(2) = (30 xx1)/300xx500/5 = 10 ` L |
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| 924. |
The vapour pressure of a solution is always less than of the pure solvent, when the temperature of both the solution and solvent are the same, with the same external pressure acting over them. Explain. |
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Answer» (i) relation between the vapour pressure of the liquid and its intermolecular forces of attraction (ii) comparison of the constituents of solution and pure solvent (iii) comparison of the intermolecular force of attraction existing in the solution and that in the existing pure solvent (iv) effect on vapour pressure |
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| 925. |
Balloons of 2 L capaicty are to be filled with hydrogen, at a pressure of 1 atm and `27^(@)C` temperature, from an 8 L cylinder containing hydrogen at 10 atm, at the same temperature . Calculate the number of balloons that can be filled. |
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Answer» (i) application of ideal gas equation (ii) calculation of volume at 1 atmospheric pressure (iii) determination of number of ballons (iv) 40 balloons |
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| 926. |
Calculate the weight of sodium bicarbonate to be dissociated to give `0.56` L of `CO_(2)` gas. |
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Answer» `{:(2NaHCO_(3) toNaCO+HO+CO),(" 2 moles"" 1 mole"):}` GMW of `NaHCO_(3)` = 84 g 1 mole of `CO_(2)` occupies 22.4 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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| 927. |
A sample of air contains nitrogen and oxygen saturated with water vapour.The total pressure is 640 mm. The vapour pressure of water vapour is 40 mm and molar ratio of nitrogen and oxygen is ` 4 : 1`. Find out the partial pressures of `N_(2) and O_(2)`. |
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Answer» Total pressure, `P_(N_(2))+P_(O_(2)) + P_(H_(2)O) = 640` mm Vapour pressure of water vapour = 40 mm `:. P_(N_(2)) + P_(O_(2)) = 640 - 40 = 600` mm Mole fraction of nitrogen = 4/5 Mole fraction of oxygen = `1/5` Partial pressure of `N_(2) = " total pressure" xx X_(N_(2)) = 600 xx 4/5 = 480` mm Partial pressure of `O_(2) = 600 xx 1/5 = 120` mm |
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| 928. |
After usage for a certain period, a cooking gas cylinder gas cylinder was considered to be empty, as no gas was coming out of it. Is the cylinder empty in its true sense? Explain what happens if the cylinder is kept in hot water or shaken vigorously. Explain by applying kinetic molecular theory. |
| Answer» No, it is not empty in its true sense. In a cooking gas cylinder, the gas was initially at a high pressure. As the gas was drawn out form the cylinder, the amount of gas in the cylinder decreases, thereby reducing the pressure inside the cylinder. The gas will come out of the cylinder as long as the pressure of the gas inside the cylinder is more than the atmospheirc pressure. When it is kept in hot water or shaken vigorously, kinetic energy of gas molecules increases and the number of collisions increases. Thus, pressure increases and gas comes out of the cylinder. | |
| 929. |
The volume of `CO_(2)` liberated at STP is _________ , on thermal decomposition of 84 g of sodium bicarbonate. |
| Answer» Correct Answer - `11.2` L | |
| 930. |
The weight of an empty china dish is 39 g and when a saturated solution of potassium nitrate is poured into it, its weight is 108 g at `50^(@) C`. After evaporating the solution to dryness, if the weight of the dish along with the crystals is 72 g them the solubility of potassium nitrate at `50^(@) C` is _________ .A. `83.9`B. ` 95.6`C. ` 91.6`D. `87.4` |
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Answer» The weight of saturated solution = 108 - 39 = 69 g The weight of `KNO_(3)` crystals = 72 - 39 = 33 g The weight of water in saturated solution = 69 - 33 = 36 g `:." Solubility of "KNO_(3) = ("Weight of" KNO_(3))/("Weight of water") xx 100 ` `= 33/36 xx 100 = 91.6" at " 50^(@)C` |
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| 931. |
A certain amount of oxygen is prepared by the thermal decomposition of potassium chlorate and is collected by downward displacement of water. The pressure of the gas collected is measured with the help of a manometer. The pressure recorded is found to be more than the pressure recorded for the same volume of oxygen cylinder containing same amount of oxygen under the same conditions. How do you account for this deviation? |
| Answer» When oxygen is prepared , it is collected over water. Since water is in equilibrium with water vapour, the oxygen gas collected is not dry oxygen gas and it is mixed with some amount of water vapour. The total pressure exerted by this moist oxygen gas is the sum of the partial pressures of oxygen and water vapour. The pressure recorded for the same volume of pure and dry oxygen gas is obviously less than previous case. The pressure of water vapour in the collected gas is called aqueous tension. | |
| 932. |
If `100 mL` of `H_(2) SO_(4)` and `100 mL` of `H_(2)O` are mixed, the mass percent of `H_(2) SO_(4)` in the resulting solution `(d_(H_(2)SO_(4)) = 0.09 g mL^(-1), d_(H_(2)O) = 1.0 mL^(-1))`A. 90B. 47.36C. 50D. 60 |
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Answer» Correct Answer - B `X_(H_(2)SO_(4)) = (100 xx 0.9)/(100 xx 0.9 + 100 xx 1) = (90)/(190) = 0.4736` % mass of `H_(2)SO_(4) = 43.36%` |
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| 933. |
Fluoro carbon polymers can be made by fluorinationg polythene. (i) `(CH_(2))_(n) + 4n Co F_(3) rarr (CF_(2))_(n) + 2 nHF + 4 nCoF_(2)` Where `n` is large integer. The `CoF_(3)` can be regenarted by the above reaction. (ii) `2CoF_(2) + F_(2) rarr 2CoF_(3)` If the `HF` formed in reactionn (i) cannot be reused, calculate the weight of `F_(2)` consumed by `1.0 g` of `(CF_(2))_(n)` produced.A. `2.0 g`B. `2.52 g`C. `1.52 g`D. `3.0 g` |
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Answer» Correct Answer - C `1.0 g (CF_(2))_(n) = ((cancel(1 "mol" (CF_(2))_(n)))/(50 cancel(n) g (CF_(2))_(n)) ((4 cancel(n) cancel("mol" CoF_(3)))/(cancel("mol"(CF_(2))_(n))))` `((1 "mol" F_(2))/(2 "mol" cancel(CoF_(3)))) ((38 g F_(2))/("mol" F_(2)))` `implies 1 xx (1)/(50) xx (4)/(1) xx (1)/(2) xx (38)/(1) implies 1.52 g F_(2)` (note for `n` cancels) |
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| 934. |
Fluoro carbon polymers can be made by fluorinationg polythene. (i) `(CH_(2))_(n) + 4n Co F_(3) rarr (CF_(2))_(n) + 2 nHF + 4 nCoF_(2)` Where `n` is large integer. The `CoF_(3)` can be regenarted by the above reaction. (ii) `2CoF_(2) + F_(2) rarr 2CoF_(3)` If `HF` can be recovered and electrolyzed to `H_(2)` and if `F_(2)`, is used for regenerating `CoF_(3)`, what is the net consuption of `F_(2)` for `1.0 g` of `(CF_(2))_(n)`.A. `1.0 g`B. `1.26 g`C. `0.76 g`D. `1.5 g` |
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Answer» Correct Answer - C If `4nCoF_(3)` yields `4nCoF_(2)`, `4nF` atoms are consumed. Saving `2 nF` atoms by recovery of the `F` from `HF` reduces the comsumption by half. Hence, `(1.52)/(2) = 0.76 g` of `F_(2)` is required. |
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| 935. |
`12.5 mL` of a solution containing `6.0 g` of a dibasic acid in `1 L` was found to be neutralized by `10 mL` a decinormal solution of `NaOH`. The molecular weight of the acid isA. 150B. 120C. 110D. 75 |
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Answer» Correct Answer - A Dibasic acid `-= NaOH` `12.5 mL` `10 mL xx 0.1 N = 1 mEq` Strength of dibasic acid ` = 6 gL^(-1)` mEq of acid = mEq of `NaOH` `12.5 xx N = 1 mEq` `(12.5 xx 6 xx 1000)/((Mw_(2))/(2) xx 1000) = 1` `:. Mw` (acid) `= 150` |
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| 936. |
Potassium dichromate `(K_(2)Cr_(2)O_(7))` is an orange coloured compound , very frequently used in laboratory as an oxidising agent as well as in a redox titration. It is generally prepared from chromite `(FeCr_(2)O_(4))` ore according to the following reactions: (1) Fusion of chromite ore with sodium carbonate in excess of air. `FeCr_(2)O_(4) + Na_(2)CO_(3) + O_(2) rarr Na_(2)CrO_(4)+Fe_(2)O_(3) +CO_(2)` (2) Acidifing filtered sodium chromate solution with sulphuric acid. `Na_(2)CrO_(4) + H_(2)SO_(4) rarr Na_(2)Cr_(2)O_(7) + Na_(2)SO_(4) + H_(2)O` (3) Treating sodium dichormate with potassium choride . `Na_(2)Cr_(2)O_(7) + KCl rarr K_(2)Cr_(2)O_(7) + NaCl` If whole of the chormite ore given in the previous question gets consumed and sufficient amount of rest of the reactants are given, then the mass of `K_(2)Cr_(2)O_(7)` obtained is:A. 14.7 gmB. 7.35 gmC. 73.5 gmD. 147 gm |
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Answer» Correct Answer - C |
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| 937. |
`500 mL` of `0.2 M NaCl` sol. Is added to `100 mL` of `0.5 M AgNO_(3)` solution resulting in the formation white precipitate of `AgCl`. How many moles and how many grams of `AgCl` are formed? Which is the limiting reagent? |
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Answer» `underset({:(500 xx 0.2),(= 100 mmol):})(NaCl) + underset({:(100 xx 0.5),(= 50 mmol):})(AgNO_(3)) rarr AgCl + NaNO_(3)` 50 mmol of `AgNO_(3)` will react with 50 mmol of `NaCl` and 50 mmol of `AgCl` will be fromed. `AgCL = 50 m "moles" = 50 xx 10^(-3) = 0.05 mol` Weight of `AgCl = 0.05 xx 143.5 = 7.175 g` Hence, `AgNO_(3)` is the limiting reagent. |
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| 938. |
When x gm carbon is burnt with y gm oxygen in a closed vessel, no residue is left behind. Which of the following statement is correct regarding the relative amounts of oxygen and carbon?A. `y/x` must lie between 1.33 and 2.67B. `y/x` must be greater than or equal to 2.67C. y/x must be less than or equal to 1.33D. y/x must be greater than or equal to 1.33 |
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Answer» Correct Answer - D |
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| 939. |
3 gm mixture of `SiO_(2)` and `Fe_(2)O_(3)` on very strong heating leaves a residue weighing 2.92gm because of conversion of `Fe_(3)O_(4)` liberating oxygen gas. What is the percentage by mass of `SiO_(2)` in original mixture?A. 0.2B. 0.8C. 0.4D. 0.6 |
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Answer» Correct Answer - A |
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| 940. |
An organometallic compound on analysis gave the following results : `C = 64.4 %, H = 5.5 %, Fe = 29.9 %`. Determine the empirical formula. |
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Answer» Correct Answer - `C_(10)H_(10)Fe` Empirical formula may be calculated as : `{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",64.4,12,(64.4)/(12)=5.4,(5.4)/(0.53)=10.2,10),("H",5.5,1,(5.5)/(1)=5.5,(5.5)/(0.53)=10.3,10),("Fe",29.9,56,(29.9)/(56)=0.53,(0.53)/(0.53)=1.0,1):}` Empirical formula of the compound `= C_(10)H_(10)Fe`. |
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| 941. |
A pure sample of cobalt chloride weighting `1.30 g` was found to contains `0.59 g` cobalt and `0.71 g` chloride on quantitative analysis. What is the percentage composition of cobalt chloride? |
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Answer» Correct Answer - A::C::D a. `1.30 g` of the compound `= 0.59 g of Co` `100 g` of compound `= (0.59 xx 100)/(1.3) = 45.4%` b. `1.30 g` of the compound `= 0.71 g` of chloride `100 g` of the compound `= (0.71 xx 100)/(1.3) = 54.6%` |
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| 942. |
`20g` of impure `NACI` smaple is added aqueous solution having excess `AGNO_(3)AgCI` precipitate is filtered [Atomic mass `Ag = 108, CI = 35.5]` . |
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Answer» `{:(AgNO_(3)(aq)+,NaCl(aq) rarr,AgCl(s)+,NaNO_(3)(s),["Precipitation reaction"]),(0.2 "moles"," "(28.7)/(143.5)"moles",,,):}` `= 0.2 "moles"` Mass of `NaCl = 0.2 xx 58.5 = 11.7 gm` `%` purity `= (11.7)/(20) xx 100` `= 58.8 %` |
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| 943. |
In `4 g` atoms of `Ag.` calculate a. Amount of `Ag.` b. Weight of one atom of `Ag`. (atoic weight of `Ag = 108`). |
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Answer» a. `4 g "atom" = 4 "mol of" Ag` `1 "mol of" Ag = 108 g` `4 "mol of" Ag = 108 xx 4 = 432 g` b. `1 "mol of" Ag = 6.023 xx 10^(23) "atoms"` `:. 6.023 xx 10^(23) "atoms" =? 108 g` `1 "atom" implies (1085)/(6.023 xx 10^(23)) g implies 17.93 xx 10^(-23) g` |
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| 944. |
6.0gm of silver of a tetrabasic acid gives 4.32 gm silver on strong heating. The molar mass of the acid is: (Ag = 108)A. 168B. 172C. 84D. 88 |
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Answer» Correct Answer - B |
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| 945. |
A compound on analysis was found to contain `C = 34.6 %, H = 3.85 % and O = 61.55%` .Calculate its empirical formula. |
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Answer» Step I : Calculation of simplest whole number ratios of the elements `{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",34.6,12,(34.6)/(12)=2.88,(2.88)/(2.88)=1,3),("H",3.85,1,(3.85)/(1)=3.85,(3.85)/(2.88)=1.337or 4//3,4),("O",61.55,16,(61.55)/(16)=3.85,(3.85)/(2.88)=1.337or 4//3,4):}` The simplest whole number ratios of the different elements are : `C : H : O : : 3 : 4: 4` Step II. Writing the empirical formula of the compound. The empirical formula of the compound `= C_(3)H_(4)O_(4)`. |
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| 946. |
A `0.2075 g` sample of an oxide of cobalt on analysis was found to contain `0.1475 g` cobalt. Calculate the empirical formula of the oxide. `(Co = 59 "amu")` |
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Answer» Weight of `O = 0.2075 - 0.1475 = 0.06 g` Mol of `Co = (0.1475)/(59) = 0.0025` Mol of `O = (0.06)/(16) = 0.003` Simplest ratio of `Co = (0.0025)/(0.0025) = 1.0` Simplest ratio of `O = (0.003)/(0.0025) = 1.5` Ratio of `Co : O = 1 : 1.5 = 2 : 3` Formula `= Co_(2) O_(3)` |
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| 947. |
The ratio of masses of oxygen and nitrogen in as particular gaseous mixture is 1 : 4. The ratio of number of their molecule is : (Atomic mass Ag = 108, Br = 80)A. `1 : 8`B. `3 : 16`C. `1 : 4`D. `7 : 32` |
| Answer» Correct Answer - D | |
| 948. |
A transition metal M forms a volatile chloride which has a vapour density of 94.8. If it contains 74.75% of chlorine the formula of the metal chloride will beA. `MCl_(2)`B. `MCl_(4)`C. `MCl_(5)`D. `MCl_(3)` |
| Answer» Correct Answer - B | |
| 949. |
`0.1653 g` aluminium reacts completely with `0.652 g` chlorine to form chloride of aluminium. a. What is the empirical formula of the compound? b. If molecular mass of the compound is 267 amu, calculate the molecular formula of the compound. |
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Answer» `0.1653 g of Al = 0.652 g` of chlorine a. `27 g of Al = (0.652)/(0.1653) xx 27` `106.95 g` of chlorine Moles of Chlorine `= (106)/(35.5) = 2.998 ~~ 3` Therefore formula `= AlCl_(3)` b. `Mw = 267 g` Empiricial formula weight of `AlCl_(3) = 27 + 3 xx 35.5` `= 133.5` `n = (Mw)/(EFw) = (267)/(133.5) = 2` Molecular formula `= 2 xx AlCl_(2) = Al_(2) Cl_(6)` |
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| 950. |
A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of `CO_(2)`. The empirical formula of the hydrocarbon isA. `C_(2)H_(4)`B. `C_(3)H_(4)`C. `C_(6)H_(5)`D. `C_(7)H_(8)` |
| Answer» Correct Answer - D | |