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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.
| 951. |
Silicon forms a compound with chlorine in which 5.6 g of silicon is combined with 21.3 og chlorine. Calculate the empirical formula of the compound (Atomic mass : `Si = 28 , Cl = 35.5`) |
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Answer» Weight of silicon in the compound `= 5.6 g` Weight of chlorine in the compound `= 21.3 g` Total weight of the compound `= 21.3 + 5.6 = 26.9 g` Percentage of silicon in the compound `= ("Weight of silicon")/("Weright of compound")xx100=((5.6g))/((26.9g))xx100=20.8` Percentage of chlorine in the compound `= ("Weight of chlorine")/("Weight of compound")xx100=((21.3g))/((26.9g))xx100=79.2` Step I. Calculation of the simplest whole number ratios of the elements `{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("Si",20.8,28,(20.8)/(28)=0.74,(0.74)/(0.74)=1,1),("Cl",29.2,35.5,(79.2)/(35.5)=2.23,(2.23)/(0.74)=3,3):}` The simplest whole number ratios of different elements are : `Si : Cl : : 1:3` Step II. Writing the empirical formula of the compound The empirical formula of the compound `= SiCl_(3)` |
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| 952. |
32 g of sulphur will react with 32 g of oxygen, even if more than 32 g of sulphur is available. Which law of chemical combination does it illustrate ? |
| Answer» The illustrates the law of constant composition because sulphur and oxygen are combined in fixed ratio by mass in sulphur dioxide `(SO_(2))` which is formed in this case. | |
| 953. |
The following data were obtained for three oxides of lead (i) 1.77 g of red oxide contains 1.61 g of lead (ii) 6.90 g of the yellow oxide contains 6.42 g of lead (iii) 1.195 g of brown oxide contains 1.035 of lead. Justify that the data is in agreement with the law of multiple proportions. |
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Answer» (i) 1.61 g of lead are combined with oxygen `= 1.77 - 1.61 = 0.16` g 1.0 g of lead are combined with oxygen `= (0.16)/(1.61) g = 0.099 g` (ii) 6.42 g of lead are combined with oxygen `= 6.9 - 6.49 = 0.48 g` 1.0 g of lead are combined with oxygen `= (0.48)/(6.42) g = 0.075 g` (iii) 1.035 g of lead are combined with oxygen `= 1.195 - 1.035 = 0.16 g` 1.0 g of lead are combined with oxygen `= (0.16)/(1.035)g = 0.1546 g` The ratio by mass of oxygen combining with 1.0 gram of lead in the three oxides is : `0.099 : 0.075 : 0.1546 or 4: 3: 6` Since it is a simple whole number ratio, if illustrates the law of multiple proportions. |
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| 954. |
Aluminium oxide contains `52.9%` aluminium and carbon dioxide contains `27.27%` carbon. Assuming the law of reciprocal proportions, calculalte the percentage of aluminium in aluminium carbide. |
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Answer» Aluuminium oxide contains : `Al = 52.9` parts, `O = 47.1` parts Carbon dioxide contains: `C = 27.27` parts, `O = 72.73` parts According to the law of reciprocal proportions, the weights of `Al` anc `C` that would combine with each other to form aluminium carbide would either be the same or simple multiple of their that combine with the same weight of oxygen in aluminium oxide and carbon dioxide. Hence, the weight of carbon that would combine with `47.1 g` of oxygen in `CO_(2) = (27.27 xx 47.1)/(72.73) = 17.62`, as also, in aluminium oxide, `52.9 g` of `Al` combine with `47.1 g` of oxygen. Thus, aluminium and carbon should be present in aluminium carbide in the ratio of `52.9 : 17.62`, i.e., `3 : 1` or a simple multiple of this ratio. Hence, in aluminium carbide, percent of `Al = (52. xx 100)/(70.52) = 75.01`. |
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| 955. |
Magnesium carbide reacts with water to give propyne gas and magnesium hydroxide. Write the balanced chamical reaction. |
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Answer» The skeleton equation is: `Mg_(2) C_(3) (s) + H_(2) O (l) rarr CH_(3) - C -= CH (g) + Mg (OH)_(2) (s)` a. Order of selection of atoms for balancing `Mg, O, H`, and `C`. b. To equalise the number of `Mg` atoms on both sides, multiply the molecule of `Mg (OH)_(2)` by 2. `Mg_(2) C_(3) (s) + H_(2) O (l) rarr CH_(3) - c -= CH (g) + 2 Mg (OH)_(2) (s)` c. To equalise the number of `O` atoms on both sides, multiply th molecule of `H_(2) O` by 4. `Mg_(2) C_(3) (s) + 4 H_(2) O (l) rarr CH_(3) - C -= CH (g) + 2 Mg (OH)_(2) (s)` It is balanced chemical equation. |
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| 956. |
Calcium carbide reacts with water to give ethyne or acetylene gas and calcium hydroxide. Write the balanced chemical equation for this reaction. |
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Answer» The skeleton equation is `CaC_(2) (s) + H_(2) O (l) rarr HC -= CH (g) + Ca (OH)_(2) (s)` a. Order of selection of atoms for balancing is `O, H`,and `C`. The number of `O` atoms on `RHS` is 2, and one `O` atom on `LHS`. So multiply `H_(2) O` with 2. `CaC_(2) (s) +2 H_(2) O (l) 0gt HC -= CH (g) + Ca (OH)_(2) (s)` It is a balanced chemical or atomic equation. |
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| 957. |
Calculate the masses of cane sugar and water required to prepare 250 g of 25 % solution of cane sugar. |
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Answer» Mass percent `=("Mass of solute")/("Mass of solution")xx100` Mass percent =25, Mass of solution = 250 g `:. 25 = ("Mass of can sugar")/((250 g))xx100` `:.` Mass of cane sugar `= (25 xx(250 g))/(100)=62.5 ,` Mass of water `= 250-62.5=187.5g`. |
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| 958. |
Which of the following statements indicates that law of multiple proportion is being followed?A. Sample of carbon dioxide taken from any source will always have carbon and oxygen namely in the ratio 1 : 2B. Carbon forms two oxides namely `CO_(2) and CO`, where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2 : 1C. When magnesium burns in oxygen, the amount of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formedD. At constant temperature and pressure 200 mL of hydrogen will combine with 100 mL of oxygen to produce 200 mL of water vapours. |
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Answer» Correct Answer - B The statement indicates that the law of multiple proportions is followed. |
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| 959. |
Law of multiple proportion is illustrated by:A. Calcium carbonate and Barium carbonateB. sodium chloride and potassium chlorideC. sulphur dioxide and sulphur trioxideD. Carbon dioxide and sulphur dioxide |
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Answer» Correct Answer - C |
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| 960. |
The density of 2 molal aqueous solution of NaOH is `1.10 "gm L"^(-1)`. Calculate the molarity of the solution. |
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Answer» 2 molal NaOH solution represents 2 moles of solute of dissolved in 1000 g of water Mass of NaOH in 2 molal solution `= (2xx 40 g)=80g` Mass of solution = Mass of solute + Mass of solvent `=(80 +1000)=1080g` Volume of solution `= ("Mass)/("Density") = ((1080 g))/((1.1 "gm L"^(-1)))=981.8mL` Molarity of solution `(M)=("Moles of solute")/("Volume of solution in litres")` `=((2mol))/((981.8//1000L))=2.04"mol L"^(-1)=2.04M`. |
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| 961. |
Statement-1 : Atomic mass of sodium is 23u Statement-2 : An atom of sodium is 23 times heavier than atom of C-12 isotopeA. Statement 1 is true, statement 2 is also true , statement 2 is the correct explanation of statement 1.B. Statement 1 is true, statement 2 is also true , statement 2 is not the correct explanation of statement 2C. Statement 1 is true, statement 2 is falseD. Statement 1 is fale, statement 2 is true |
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Answer» Correct Answer - C 1 atom of sodium is 23 times heavier than 1/12 of an atom of C-12 istope. |
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| 962. |
The density of 3 molal solution of NaOH is 1.110g `mL^(-1)`. Calculate the molarity of the solution. |
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Answer» 3 molal NaOH solution represents 3 moles of solute dissolved in 1000 g of water Mass of the solution = (Mass of the solute)+(Mass of the solvent) `=(3 mol xx 4 g mol^(-1))+(1000g)` `=120g + 1000g = 1120 g` Volume of the solution `= ("Mass")/("Density")=(1120 g)/((1.110 "g mL"^(-1)))=1009 mL` Molarity of the solution `(M)=("Moles of the solution")/("Volume of the solution in litres")=((3mol))/((1009//1000L))=2.97"mol L"^(-1)=2.97M` |
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| 963. |
An element exist in nature in two isotopic forms: `X^(30)(90%) and X^(32)(10%)`. What is the average atomic mass of element? |
| Answer» Av. Atomic mass `=(sum(%"abundance"xx"atomic mass"))/(100)=(90xx30+10xx32)/(100)=30.2` | |
| 964. |
In coal, pyrites `(FeS_(2))` is present as a pollution-causing impurity, which is removed by combustion. `2FeS_(2) + 5O_(2) rarr 4SO_(2) + 2FeO`. What mass of `H_(2)SO_(4)` can be prepared from `3.0 g` of `Cu_(2) S` if each atom of `S` in `Cu_(2)S` is converted into 1 molecule of `H_(2) SO_(4)`?A. `1.85 g`B. `68.62 g`C. `3.85 g`D. `4.85 g` |
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Answer» Correct Answer - A `Cu_(2) S rarr H_(2) SO_(4)` Weight of `H_(2) SO_(4) = (3.0 g Cu_(2) S)` `((1 "mol" Cu_(2) S)/(159 g Cu_(2) S))((1 "mol" S)/("mol" S))` `((98 g H_(2) SO_(4))/("mol"H_(2)SO_(4)))` `= (3 xx 1 xx 1 xx 98)/(159) = 1.85 g H_(2) SO_(4)` |
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| 965. |
In coal, pyrites `(FeS_(2))` is present as a pollution-causing impurity, which is removed by combustion. `2FeS_(2) + 5O_(2) rarr 4SO_(2) + 2FeO`. A process designed to remove orgainc sulphur from coal prior to combustion involves the reaction. `X - S - Y + 2NaOH rarr X - O - Y + Na_(2)S + H_(2) O` `CaCO_(3) rarr CaO + CO_(2)` `Na_(2)S + CO_(2) + H_(2)O rarr Na_(2) CO_(3) + H_(2) S` `CaO + H_(2)O rarr Ca (OH)_(2)` `Na_(2) CO_(3) + Ca(OH)_(2) rarr CaCO_(3) + 2NaOH` In the processing in 320 metric tons of a coal having 1.0% sulphur content, how much limestone `(CaCO_(3))` must be edecomposed to provied enough `Ca (OH)_(2)` to regenerate the `NaOH` used in the original leaching step?A. 2.0 metric tonB. 4.0 metric tonC. 8.0 metric tonD. 10.0 metric ton |
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Answer» Correct Answer - D Weight of `CaCO_(3) = (320 xx 10^(6) g "coal")` `((1.0 g S)/(100 g "coal"))((1 "mol"S)/(32 g S))` `((1 "mol" CaCO_(3))/("mol" S))((100 g CaCO_(3))/("mol"CaCO_(3)))` `= (320 xx 10^(6) xx 1 xx 1 xx 1 xx 100)/(100 xx 32)` `= 10 xx 10^(6) g CaCO_(3)` = 10 metric ton `CaCO_(3)` |
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| 966. |
What volume of 90% alcohol by weight `(d = 0.8 g mL^(-1))` must be used to prepared `80 mL` of 10% alcohol by weight `(d = 0.9 g mL^(-1))` |
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Answer» Correct Answer - A Use `M = (% "by weight" xx 10 xx d)/(Mw_(2))` `M_(1) V_(1) = M_(2) V_(2)` `(90 xx cancel(10) xx 0.8)/(cancel(46)) xx V = (10 xx cancel(10) xx 0.9)/(46) xx 80` `V = (10 xx 0.9 xx 80)/(90 xx 0.8) = 10 mL` |
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| 967. |
In coal, pyrites `(FeS_(2))` is present as a pollution-causing impurity, which is removed by combustion. `2FeS_(2) + 5O_(2) rarr 4SO_(2) + 2FeO`. What volume of `3.0 M KOH` would be required to react with the `SO_(2)` produced inA. `2.77 L`B. `5.54 L`C. `1.38 L`D. `8.31 L` |
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Answer» Correct Answer - B `2KOH + SO_(2) rarr K_(2) SO_(3) + H_(2) O` Moles of `KOH = 2 xx` moles of `SO_(2) = 2 xx 8.32` `M xx V_(L) =` Moles of `SO_(2)` `3 xx V_(L) = 2 xx 8.32` `V_(L) = 5.54 L` |
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| 968. |
In coal, pyrites `(FeS_(2))` is present as a pollution-causing impurity, which is removed by combustion. `2FeS_(2) + 5O_(2) rarr 4SO_(2) + 2FeO`. Calculate the moles of `SO_(2)` produced by burning 1.0metric ton `(10^(3) kg)` of coal containing 0.05% by mass of pyrites impurity?A. `8.32 mol`B. `4.16 mol`C. `12.48 mol`D. `2.08 mol` |
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Answer» Correct Answer - A Mol of `SO_(2) = (1.0 xx 10^(6) g "coal")` `((0.05 g Fe S_(2))/(100 g "coal"))((1 "mol" g FeS_(2))/(120 g "Coal"))` `((4 "mol" SO_(2))/(2 "mol" FeS_(2)))` `= (10^(6) xx 0.05 xx 1 xx 4)/(100 xx 120 xx 2) = 8.32 "mol" SO_(2)` |
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| 969. |
`50 mL` of `1 M HCl, 100 mL` of `0.5 M HNO_(3)`, and `x mL` of `5 M H_(2)SO_(4)` are mixed together and the total volume is made upto `1.0 L` with water. `100 mL` of this solution exactly neutralises `10 mL` of `M//3 Al_(2) (CO_(3))_(3)`. Calculate the value of `x`. |
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Answer» Correct Answer - A Total mEq of acid `= 50 xx 1 + 100 xx 0.5 + x xx 5 xx 2` (`n` factor ) `= (100 + 10x) mEq` `= ((100 + 10 x))/(100 mL) N` `N_(1) V_(1)` (Acid) `= N_(1) V_(1) [Al_(2) (CO_(3)^(2-))_(3)]` (Total charge = 6) `(n = 6)` `:. ((100 + 10 x))/(100 mL) N xx 100 mL = 10 mL xx (1)/(3) xx 6` `(100 + 10x) = 200` `:. x = 10 mL` |
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| 970. |
`15 mL 1 N H_(2) SO_(4)`, `25 mL` of `4 N HNO_(3)`, and `20 mL` of `X M HCl` were mixed and made up to `1000 mL`. Prepared by dissolving `4.725 g` of pure `Ba(OH)_(2). 8H_(2) O` in water made up to 0.25 litre. What is the molarity of `HCl` solution (i.e. find `X`) |
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Answer» `15 mL` of `1 M H_(2) SO_(4) + 25 mL of 4 M HNO_(3) + 20 mL of X M HCl` `:. N_(1) V_(1) + N_(2) + V_(2) + N_(3) V_(3) = N_(4) V_(4)` `(V_(4) = 1000 mL)` `15 xx 2 + 25 xx 4 + 20 X = N_(4) xx 1000` `:. N_(4) = ((130 + 20 X)/(1000))` mEq of mixture of acid = mEq of `Ba(OH)_(2)` `Mw of Ba (OH)_(2). 8H_(2) O = 137.4 + 34 + 18 xx 8 = 315.4` `Ew = (315)/(2) = 157.7 g` `N of Ba(OH)_(2) .8H_(2) O = (W_(2) xx 1000)/(Ew_(2) xx V_(sol) ("in" mL))` `= (4.725 xx 1000)/(157.7 xx 250)` `0.1198 N ~~ 0.12 N` mEq of acid mix = mEq of `Ba(OH)_(2)` `20 xx N_(4) = 26 xx 0.12` `N_(4) = (26 xx 0.12)/(20) = 0.156 N` `implies (130 + 20 X)/(1000) = 0.156` `:. X = (0.156 xx 1000 - 130)/(20) = 1.3` `N` or `M HCl = 1.3` |
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| 971. |
In a binary solution, the component present in smaller proportion is called solute while the one in excess is known as solvent. The solution containing 1 mole of the solute in 1L of solution is known as one molar solution while the solution in which 1 mole of solute is dissolved in 1 kg of the solvent is called one molal solution. The ratio of the no. of moles of a particular component to the total no. of moles in the solution is known as its mole fraciton The unit of molarity areA. MolesB. Moles/kgC. Moles/litreD. grams/litre |
| Answer» Correct Answer - C | |
| 972. |
`1 L` of `0.1 M NaOH, 1 L` of `0.2 M KOH`, and `2 L` of `0.05 M Ba (OH)_(2)` are mixed togther. What is the final concentration of the solution. |
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Answer» Total volume `(V_(4)) = 1 + 1 + 2 = 4 L` `n` factor of `NaOH KOH` is 1, while for `Ba(OH)_(2)` is 2. `N_(1) V_(1) + N_(2) V_(2) + N_(3) V_(3) = N_(4) V_(4)` `0.1 xx 1 + 0.2 xx 1 + 0.05 xx 2 xx 2 = N_(4) xx 4` `0.1 + 0.2 + 0.2 = N_(4) xx 4` `N_(4) = (0.4)/(4) = 0.1 N` Hence, final concentration of solution `= 0.1 N` |
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| 973. |
`100 mL` of `0.1 M HCl + 100 mL` of `0.2 M H_(2)SO_(4) + 100 mL` of `0.1 M HNO_(3)` are mixed togther. a. What is the final conecntration of the solution. b. What would be the final concentration of the solution. If the solution is made to `1 L` by adding `H_(2) O`? c. What would be the final concentration of the solution if `700 mL` of `H_(2)O` is added to the solution? |
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Answer» a. `N_(1) V_(1) + N_(2) V_(2) + N_(3) V_(3) = N_(4) V_(4)` `(V_(4) = V_(1) + V_(2) V_(3) = 100 + 100 + 100 = 300 mL)` `n` factor for `HCl` and `HNO_(3) = 1` `n` factor for `H_(2) SO_(4) = 2` In such questions always convert molarity to normality, since 1 grams equivalent of one reactant is always equal to 1 gram equivalent of another reactant. `100 xx 0.1 + 100 xx 0.2 + 2 100 xx 0.1 = N_(4) xx 300 + 10 + 40 + 10 = N_(4) xx 300` `:. N_(4) = (60)/(300) = (1)/(5) = 0.2 N` Hence, final concentration of solution `= 0.2 N` b. Final volume `= 1 L = 1000 mL` `:. 100 xx 0.1 + 100 xx 0.2 xx 2 + 100 xx 0.1 = N_(4) xx 1000` `N_(4) = (60)/(1000) = 0.06 N` Hence, final concentration of solution `= 0.06 N` c. Final volume `= 300 + 700 = 1000 mL` `:. N_(4) = (60)/(1000) = 0.06` Hence, final concentration of solution `= 0.06 N` |
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| 974. |
Two students performed the same experiment separately and each one of them recorded two readings of mass which are given below. However, the correct reading is 5.0 g. On the basis of the data, mark the correct option `{:("Student",:underset("(i) (ii)")("Readings")),("A","5.01 4.99"),("B","5.05 4.95"):}`A. Results of both the students are neither accurate nor preciseB. Results of student A are both accurate and preciseC. Results of student B are neither precise nor accurateD. Results of student B are both precise and accurate |
| Answer» Correct Answer - B | |
| 975. |
In the preceding problem, if the compound contains `12.8%` nitrogen and `9.8%` sulphur how many nitrogen and sulphur atoms are present per atomof sodium?A. `2` and `1`B. `1` and `3`C. `1` and `2`D. `3` and `1` |
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Answer» Correct Answer - D `{:(,,"mole","atom"),(,%N=12.8,12.8//14,(12.8)/(14)xxN_(A)),(,%S=9.8,9.8//32,(9.8)/(14)xxN_(A)),(,%Na=7,7//23,(7)/(23)xxN_(A)):}` `because (7)/(23)xxN_(A)` atom of Na contain `=(12.8)/(14)xxN_(A)` of `N` `therefore 1` atom of `Na` contain `=3` atom of `N` `therefore (7)/(23)xxN_(A)` atom of Na contain `=(9.8)/(32)xxN_(A)` of `S` `therefore 1` atom of Na contain `=(9.8)/(32)xx(23)/(7) =1` atom |
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| 976. |
Calculate the molecular mass of the following: a. `H_(2)O` b. `CO_(2)` c. `CH_(4)` |
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Answer» Molecular mass of `Na_(2)SO_(4)= 2 xx` Atomic mass of `Na +` Atomic mass of `S + 4 xx` Atomic mass of O `= 2 xx 23 + 32 + 4 xx 16 = 46 + 32 + 64 = 142` u The percentage of different present can be calculated as : Percentage of sodium `(Na)=("Mass of Na")/("Molecular mass of "Na_(2)SO_(4))xx100=(46)/(142)xx100=32.39%` Percentage of sulphur `(S)=("Mass of S")/("Molecular mass of "Na_(2)SO_(4))xx100=(32)/(142)xx100=22.54%` Percentage of oxygen `(O)=("Mass of O")/("Molecular mass of "Na_(2)SO_(4))xx100=(64)/(142)xx100=45.07%`. |
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| 977. |
The sodium salt of methyi orange has `7%` sodium. What is the minimum molecular weight of the compound? :A. `420`B. `375`C. `329`D. `295` |
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Answer» Correct Answer - C `{:(1grarr=(100)/(7)xx23),(23grarr=329):}` |
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| 978. |
In the section `:Na_(2)S_(2)O_(3)+4Cl_(2)+5H_(2)O rarr Na_(2)SO_(4)+H_(2)SO_(4)+8HCl`. the equivalent weight of `Na_(2)S_(2)O_(3)` will be : `(M=` molecular weight of `Na_(2)S_(2)O_(3)`)A. `M//4`B. `M//8`C. `M//1`D. `M//2` |
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Answer» Correct Answer - 2 `Na_(2)overset(+2)(S)_(2)o_(3)rarr Na_(2)overset(+6)(S)O_(4)` the total change in oxidation number `=4xx2=8` `:. " "E_(Na_(2)S_(2)O_(3))=(mol.wt.)/(V.f)=(M)/(8)` |
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| 979. |
The number of moles of `KNnO_(4)` that will be needed to react with one mole of sulphite ion in acidic solution is :A. `4//5`B. `2//5`C. 1D. `3//5` |
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Answer» Correct Answer - B The ionic equation for the reaction is : `2MnO_(4)^(-)+6H^(+)+5SO_(3)^(2-)rarr2Mn^(2+)+5SO_(4)^(2-)+3H_(2)O` 5 mole of `SO_(3)^(2-)` ions are oxidised by `MnO_(4)^(-)` ions = 2 mol 1 mole of `SO_(3)^(2-)` ions is oxidised by `MnO_(4)^(-)` ions = 2/5 mol. |
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| 980. |
How many moles of lead (II) chloride are formed from a reaction between 6.5 g of PbO and 3.2 of HCl ?A. 0.011B. 0.029C. 0.044D. 0.333 |
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Answer» Correct Answer - B `underset("1 mol")(PbO)+underset("2 mol")(2HCl)rarrunderset("1 mol")(PbCl_(2))+underset("2 mol")(2H_(2)O)` `underset("0.029 mol")((6.5)/(224))"mol "underset("0.087 mol")((3.2)/(36.5))"mol"` PbO is limiting reaction `:.` No. of moles of `PbCl_(2) = 0.029` mol. |
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| 981. |
The red colour of blood is due to haemoglobin. It contain `0.335` percent of iron. Four atoms of iron are present in one molecule of haemoglobin. What is the molecular mass of haemoglobin ? (Given that the atomic mass of Fe = 55.84) |
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Answer» 1 atom of iron will correspond to atomic mass of iron `= 55.84` parts 4 atoms of iron will correspond to mass of iron `= 55.84 xx 4` `=223.36` parts Now, `0.335` part of iron will correspond to mass of haemoglobin = 100 parts `:. 223.36` parts of iron will correspond to mass of haemoglobin `= (100 "parts") xx ((223.36 "parts"))/((0.335 "part"))` `= 66675` parts Molecular mass of haemoglobin `= 66675 u`. |
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| 982. |
When `C_(2)H_(4)` is burnt in air, carbon dioxide and water are formed. If on combustion of `C_(2)H_(4),2` moles of `CO_(2)` is produced, calculate the number of drops of water produced, calculate the number of drops of water produced along with this quantity of `CO_(2)` if each drop contains `6.022 xx 10^(21)` water molecules. |
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Answer» Correct Answer - 200 |
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| 983. |
The ratio of mass percent of C and H of an organic compound `(C_(X)H_(Y)O_(Z))` is 6 : 1 If one molecule of the above compound `(C_(X)H_(Y)O_(Z))`contains half as much oxygen as required to burn one molecule of compound `C_(X)H_(Y)` completely to `CO_(2)` and `H_(2)O`. The empirical formula of compound `C_(X)H_(Y)O_(Z)` isA. `C_(2)H_(4)O`B. `C_(3)H_(4)O_(2)`C. `C_(2)H_(4)O_(3)`D. `C_(3)H_(6)O_(3)` |
| Answer» Correct Answer - C | |
| 984. |
The shape of Tobacco Mosaic Virus (TMC) is cylindrical, having length and diameter `3000Å` and `170Å`, respectively. The density of the virus is `0.08gm//ml`. The molecular weight of TMC is:A. 3.28B. `5.44 xx 10^(-24)`C. `5.44 xx 10^(-18)`D. `3.28 xx 10^(6)` |
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Answer» Correct Answer - D |
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| 985. |
1 game of a carbonate `(M_(2)CO_(3))` in g `mol^(-1)` is :-A. `1186`B. `84.3`C. `118.6`D. `11.86` |
| Answer» Correct Answer - B | |
| 986. |
The density of HCl equal to `1.17g//mL`. The molarity of the solution will be :A. `36.5`B. `18.25`C. `19.17`D. `4.65` |
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Answer» Correct Answer - C |
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| 987. |
`1.11` gm of `CaCl_(2)` is added to water forming 500 ml solution. 20 ml of this solution is taken and diluted 10 folds. Find moles of `Cl^(-)` ions in 2 ml of diluted solution :A. `8xx10^(-6)`B. `4xx10^(-6)`C. `12xx10^(-8)`D. `5xx10^(-6)` |
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Answer» Correct Answer - A |
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| 988. |
A metal carbonate decomposes according to the following reaction `M_(2)CO_(3)` (s) `to` `M_(2)O`(s) + `CO_(2)`(g) Percentage loss in mass on complete decomposition of `M_(2)CO_(3)`(s) (Atomic mass of M= 102 )A. `100/3 %`B. `50/3 %`C. `25/3 %`D. `15%` |
| Answer» Correct Answer - B | |
| 989. |
An oleum sample is labelled as 113.5`%`. Identity the incorrect statementA. The amount of free `SO_(2)` in 50g oleum sample is 30 gB. The amount of `H_(2)SO_(4)` in 50g oleum sample is 30gC. The new labelling of oleum sample when 8 g water is added in 100 g original oleum sample is `(100+(137.5)/(27))%`D. In the original 50g oleum sample when 6.75 g is added then 56.75g `H_(2)SO_(4)` is produced. |
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Answer» Correct Answer - B |
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| 990. |
A mixture of magnesium chloride and magnesium sulphate is known to contain 0.6 moles of chloride ion and 0.2 moles of sulphate ions. For quantitative estimation of Mg, above mixture is treated with set of reagents to form `Mg_(2)P_(2)O_(7)` produced is:A. 0.5B. 0.25C. 0.8D. 0.7 |
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Answer» Correct Answer - B |
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| 991. |
A solution of magnesium chloride that is `5.10%` magnesium by mass has a density `1.17g//mL`. How many moles of `Cl^(-)` ions are in 300 mL of the solution?A. `0.377`B. `0.627`C. `0.737`D. `1.49` |
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Answer» Correct Answer - D |
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| 992. |
What mass of magnesium hydroxide is required to neutralize 125 mL of `0.136` M hydrochloric acid solution? `{:("Substance","Molar Mass"),("MG"(OH)_(2),"58.33 gmol"^(-1)):}`A. `0.248g`B. `0.496g`C. `0.992g`D. `1.98g` |
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Answer» Correct Answer - B |
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| 993. |
124 u of `P_4` will contains :A. 4 `N_A` atoms of phosphorusB. 4 atoms of phosphorusC. 1 molecule of phosphorusD. `N_A` molecules of phosphorus |
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Answer» Correct Answer - B::C |
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| 994. |
A metal carbonate decomposes according to following reaction `M_(2)CO_(3)(s) rArr M_(2)O_(s) + CO_(2)(g)` Percentage loss in mass on complete decomposition of `M_(2)CO_(3)(s)` (Atomic mass of M = 102) is:A. `(100/3)%`B. `(50/3)%`C. `(25/3)%D. 0.15 |
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Answer» Correct Answer - B |
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| 995. |
Which compound contains the highest percentage of magnesium by mass?A. `MgNH_(4)PO_(4)`B. `Mg(HPO_(4))_(2)`C. `Mg_(2)P_(4)O_(7)`D. `Mg_(3)(PO_(4))_(2)` |
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Answer» Correct Answer - D |
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| 996. |
What is the percent by mass of nitrogen in ammonium carbonate, `(NH_(2)^(4)CO_(3))`?A. 0.1453B. 0.2783C. 0.2916D. 0.3334 |
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Answer» Correct Answer - C |
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| 997. |
Which compound contains the highest percentage of nitrogen by mass?A. `NH_(2)OH` (M = 33)B. `NH_(4)NO_(2)` (M = 64.1)C. `N_(2)O_(3)` (M = 76.0)D. `NH_(4)NH_(2)CO_(2)` (M = 78.1) |
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Answer» Correct Answer - B |
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| 998. |
Lithium metal reacts with nitrogen gas to produce a white solid Lithium nitride `(Li_N)` according to the reaction : `6 Li(s)+N_2 (s)to2Li_3N(s)` if 8.4 g of Li is taken initially with excess of `N_2(g)`, then (Atomic weight of Li=7)A. Volume of nitrogen gas consumed at STP is 4.54 LB. Total mass of the product obtained is 14 gC. Total number of atoms obtained in the prouduct is `1.6 N_A`D. Lithium and nitrogen combine in ratio 3:2 : by mass. |
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Answer» Correct Answer - A::B::C::D |
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| 999. |
`0.250g` of an element M, reacts with excess fluorine to produce `0.547g` of the hexafluoride `MF_(6)`. What is the elemejnt : `[Cr = 52, Mo = 96, S = 32, Te = 127.6]` .A. CrB. MOC. SD. Te |
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Answer» Correct Answer - B `M + 6F rarr MF_(6)` Mole of M = Mole of `MF_(6)` `("wt")/("Mole wt") = ("wt")/("Mol wt")` `(.25)/(x) = (.547)/(x + 19 xx 6)` `28.5 + .25x = .547x` `28.5 = .297 x rArr x = 95.959` so element is `= Mo` . |
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| 1000. |
Calculate the number of molecules present in one drop of `H_(2) O` whose mass is `0.01 g` b. Calculate the number of molecules leaving the liquid suface per second, if the same drop of water evaporates in one hour. |
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Answer» a. `0.01 g of H_(2) O = (0.01)/(18) mol` `= (0.01)/(18) xx 6.022 xx 10^(23)` molecules b. Number of moleucles leaving per second `= (3.345 xx 10^(20))/(60 x 60) = 9.293 xx 10^(16)` molecules |
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