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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1001. |
Calculate the total number of electrons present in `3.2 g` of oxygen gas. |
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Answer» (Atomic numbe of `O = 8`, number of electrons in `O_(2) = 16`) `3.2 g of O_(2) = (3.2)/(932) = 0.1 mol` `0.1 xx 6.022 xx 10^(23)` molecules `= 6.002 xx 10^(22)` molecules `= 6.022 xx 10^(22) xx 16` electrons `= 9.635 xx 10^(23)` electrons |
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| 1002. |
A solution contains 20 g of sodium chloride in `95 cm^(3)` solution. The density of solution is `1.25 "g cm"^(-3)`. What is the mass percent of NaCl ? |
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Answer» Correct Answer - `16.84 %` Density of NaCl solution `= 1.25 "g cm"^(-3)` , Volume of solution `= 95 cm^(3)` Mass of solution = volume `xx` density `= ("95 cm"^(3))xx("1.25 g cm"^(-3))=118.75`g Mass percent of NaCl `= ("Mass of NaCl")/("Mass of solution")xx100` `=((20g))/((118.75g))xx100=16.84%`. |
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| 1003. |
1300 gm of a solution of urea having molarity equal to 5 m is kept in a large bucket is kept under a tap through which a 1 m urea solution is flowing . Assuming a constant rate of flow of urea solution which is equal to 0.5 gm/sec answer the question that follow. Time taken (in seconds) for concentration in the bucket to reach 3 m.A. 2000 secB. 2120 secC. 1980 secD. 400 sec |
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Answer» Correct Answer - B |
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| 1004. |
1300 gm of a solution of urea having molarity equal to 5 m is kept in a large bucket is kept under a tap through which a 1 m urea solution is flowing . Assuming a constant rate of flow of urea solution which is equal to 0.5 gm/sec answer the question that follow. Total amount of solution (in gm) finally present in bucket when solution present in bucket have concentration of urea equal to 3 m.A. 2360 gmB. 2000 gmC. 2200 gmD. 2480 gm |
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Answer» Correct Answer - A |
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| 1005. |
1300 gm of a solution of urea having molarity equal to 5 m is kept in a large bucket is kept under a tap through which a 1 m urea solution is flowing . Assuming a constant rate of flow of urea solution which is equal to 0.5 gm/sec answer the question that follow. Calulate approximate amount of water present in the bucket at time t=2000 sec.A. 2000 gmB. 2200 gmC. 2100 gmD. 1943 gm |
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Answer» Correct Answer - D |
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| 1006. |
Molality : It is defined as the moles of the solute pressent in 1 kg of the solvent . It is denoted by m. Molality(m)`=("Number of moles of solute")/("Number of kilograms of the solvent")` let `w_(A)` grams of the solute of molecular mass `m_(A)` be present in `w_(B)` grams of the solvent, then: Molality(m) =`(w_(A))/(m_(A)xxw_(B))xx1000` Relation between mole fraction and molality: `X_(A)=(n)/(N+n) "and" X_(B)=(N)/(N+n)` `(X_(A))/(X_(B))=(n)/(N)=("Moles of solute")/("Moles of solvent")=(w_(A)xxm_(B))/(w_(B)xxm_(A))` `(X_(A)xx1000)/(X_(B)xxm_(B))=(w_(A)xx1000)/(w_(B)xxm_(A))= m or (X_(A)xx1000)/((1-X_(A))m_(B))=m` The molality of 1 litre solution with y% by (w/v) pf `CaCO_(3)` is 2 . The weight of the solvent present in the solution is 900g , then value of y is : [Atomic weight : Ca=40, C=12 , O=16]A. 9B. 18C. 27D. 36 |
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Answer» Correct Answer - B |
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| 1007. |
Molality : It is defined as the moles of the solute pressent in 1 kg of the solvent . It is denoted by m. Molality(m)`=("Number of moles of solute")/("Number of kilograms of the solvent")` let `w_(A)` grams of the solute of molecular mass `m_(A)` be present in `w_(B)` grams of the solvent, then: Molality(m) =`(w_(A))/(m_(A)xxw_(B))xx1000` Relation between mole fraction and molality: `X_(A)=(n)/(N+n) "and" X_(B)=(N)/(N+n)` `(X_(A))/(X_(B))=(n)/(N)=("Moles of solute")/("Moles of solvent")=(w_(A)xxm_(B))/(w_(B)xxm_(A))` `(X_(A)xx1000)/(X_(B)xxm_(B))=(w_(A)xx1000)/(w_(B)xxm_(A))= m or (X_(A)xx1000)/((1-X_(A))m_(B))=m` What is the quantity of water that should be added to 16 g methonal to make the mole fraction of methonal as 0.25?A. 27 gB. 12 gC. 18 gD. 36 g |
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Answer» Correct Answer - A |
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| 1008. |
"Suvarnabhasm", an ayurvedic drug, is found to contain `400 "ppm"` of colloidal gold. Mass `%` of gold `("atomic mass of Au" = 197)` will be :A. `0.040 %`B. `7.88 %`C. `0.0788 %`D. `4xx10^(-4) %` |
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Answer» Correct Answer - A ` pp m = ("moles of solute")/("mass of solution")xx10^(6)` `(400)/(100)xx100=("moles of solute")/("mass of solution")xx100` Mass `%=0.04` |
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| 1009. |
How many moles of gold are present in `49.25 g` of gold rod ? (Atomic mass of gold = 197) |
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Answer» Correct Answer - 0.25 mol Gram atomic mass of gold `= 197.0 g` 197.0 g of gold have mass = 1 gramk mol 49.25 g of gold have mass `= ((49.25g))/((197.0g))xx("1 gram mol")=0.25` gram mol = 0.25 mol |
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| 1010. |
2 moles of nitrogen atoms at STP occupy a volume of:A. 11.35 LB. 45.4 LC. 22.7 LD. 5.6L |
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Answer» Correct Answer - C |
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| 1011. |
The atomic weight for a triatomic gas is a. The correct formula for the number of moles of gas in its w g is:A. `(3w)/(a)`B. `(w)/(3a)`C. 3waD. `(a)/(3w)` |
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Answer» Correct Answer - B |
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| 1012. |
If the mass of 0.25 moles of an element Xis 2.25g, the mass of one atom of X is about:A. `1.5 xx 10^(-24)`gB. `2.5 xx 10^(-23)`gC. `1.5 xx 10^(-23)`gD. `2.5 xx 10^(-24)`g |
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Answer» Correct Answer - C |
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| 1013. |
if two moles of an ideal at `546K` occupies a volume of `44.8` litres, the pressure must be:A. 2 atmB. 3 atmC. 4 atmD. 1 atm |
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Answer» Correct Answer - A |
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| 1014. |
A `15mL` sample of `0.20 M" " MgCl_(2)` is added to `45ML` of `0.40 M AlCl_(3)` What is the molarity of `Cl` ions in the final solutionA. `1.0M`B. `0.60M`C. `0.35M`D. `0.30M` |
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Answer» Correct Answer - A `"Molarity of" Cl^(-)` `=(M_(1)V_(1)+M_(2)V_(2))/("Total vol")` `=(15 xx .2 xx 2 + 45 xx .45 xx 3)/(15 + 45) = (60)/(60) = 1M` . |
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| 1015. |
Calculate the molarity `(M)` and normality `(N)` of a solution of oxalic acid `[(COOH)_(2) . 2H_(2) O]` containing `12.6 g` of the acid in `500 mL` of the solution. |
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Answer» Molar mass of `(COOH)_(2). 2H_(2)O = 2(12 + 32 + 1) + 2 (18)` `= 90 + 36 = 126 g` Equivalent mass of `(COOH)_(2). 2H_(2)O = ("Molar mass")/(n)` `= (126)/(2) = 63 g` (`n = 2`, since two `H^(o+)` inos are replaceable, i.e., dibasic acid) `M = (W_(2) xx 1000)/(Mw_(2) xx V_(sol)("in" mL)) = (12.6 xx 1000)/(126 xx 500) = 0.02 M` `N = (W_(2) xx 1000)/(Ew_(2) xx V_(sol)("in" mL)) = (12.6 xx 1000)/(63 xx 500) = 0.4 N` (Alternatively, `N = n xx M = 2 xx 0.2 = 0.4 N)` |
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| 1016. |
The density of 5% aqueous `MgCl_(2)` solution is `1.043 g mL^(-1)`. What is the molarity and molaltiy of the solution? What is the molality of `Cl^(ɵ)` ions? `(Mg = 24 "amu")` |
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Answer» `Mw of MgCl_(2) = 24 + 35.5 xx 2 = 95 g` `M = (% "by weight" xx 10 xx d)/(Mw_(2)) = (5 xx 10 xx 1.043)/(95) = 0.548`. Weight of solvent `(W_(1)) = 100 - 5 = 95 g` `m = (W_(2) xx 1000)/(Mw_(2) xx W_(1)) = (5 xx 1000)/(95 xx 95) = 0.554` `{:(MgCl_(2)rarr, Mg^(2+), +, 2Cl^(ɵ)),(1m, 1m, 2m),(0.554m, 0.554 m, 2 xx 0.554):}` |
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| 1017. |
If `4 g NaOH` are dissolved in `100 mL` of aqueous solution, what will be the differnce in its normality and molarity? |
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Answer» Correct Answer - A `(Mw of NaOH = 23 + 16 + 1 = 40)` `M = (W_(2) xx 1000)/(Mw_(2) xx V_(sol)) = (4 xx 1000)/(40 xx 100) = 1` Since `NaOH` is monobasic acid it normality and molarity will be same) |
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| 1018. |
The oxides of a certain (hypothetical) element contain 27.28%, 42.86% and 52.94% oxygen. What is the ratio of the valencies of the element in the 3 oxides?A. `2:3:4`B. `1:3:4`C. `1:2:4`D. `1:2:3` |
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Answer» Correct Answer - D |
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| 1019. |
Calculate the number of oxygen atoms requried to combine with `7.0 g` of `N_(2)` to form `N_(2) O_(3)` if 82% of `N_(2)` is converted into products. `N_(2) + (3)/(2) O_(2) rarr N_(2) O_(3)`A. `3.24 xx 10^(23)`B. `3.6 xx 10^(23)`C. `18 xx 10^(23)`D. `6.02 xx 10^(23)` |
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Answer» Correct Answer - B `2N_(2) + 3O_(2) rarr 2N_(2) O_(3)`, (%yield = 80%) Initial mol of `N_(2) = (7.0)/(28) = 0.25 "mol"` Mole of `N_(2)` converted `= 0.25 xx (80)/(100) = 0.2` `2 "mo" N_(2) = 3 "mol" O_(2)` `0.2 "mol" N_(2) = 0.3 "mol" O_(2) (1 "mol" O_(2) = 2` oxygen atom) `= 2 xx 0.3 "mol" O "atom"` `= 2 xx 0.3 xx 6.02 xx 10^(23) = 3.6 xx 10^(23)` |
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| 1020. |
Calculate the number of oxygen atoms requried to combine with `7.0 g` of `N_(2)` to form `N_(2) O_(3)` if 82% of `N_(2)` is converted into products. `N_(2) + (3)/(2) O_(2) rarr N_(2) O_(3)`A. `4.5 xx 10^(23)`B. `3.6 xx 10^(23)`C. 1.8 xx 10^(23)`D. `7.2 xx 10^(23)` |
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Answer» Correct Answer - B |
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| 1021. |
Silver is removed from the solutions of its salts with metallic zinc, according to the reaction `Zn + 2 Ag^(o+) rarr Zn^(2+) + 2Ag`. A `65.4 g` piece of `Zn` is put into a `100 L` vat containing `3.25 g` `Ag^(o+)` per litre. How amny moles of reactant remained unreacted? |
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Answer» Correct Answer - A Total Weight of `Ag^(o+) = ((3.24 g)/(1L)) (100 L) = 324 g` Mol of `Ag^(o+) = (324)/(108 g Ag) = 3 "mol"` Mol of `Zn = (65.4)/(65.4 g Zn) = 1 "mol" = 2 "mol of" Ag^(o+)` Since the mole ratio `Ag^(o+)//Zn` present `(3//1)` exceeds the mole ratio required `(2//1)` in the reaction, the `Ag^(o+)` is in excess, the `Zn` is completely consumed. Moles of `Ag^(o+)` consumed = 2 mol Mole of `Ag^(o+)` left `= 3 - 2 = 1` mol of `Ag^(o+)` excess |
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| 1022. |
A mixture of a mol of `C_(3) H_(8)` and `b` mol of `C_(2) H_(4)` was kept is a container of `V L` exerts a pressure of 4.93 atm at temperature `T`. Mixture was burnt in presence of `O_(2)` to convert `C_(3) H_(8)` and `C_(2) H_(4)` into `CO_(2)` in the container at the same temperature. The pressure of gases after the reaction and attaining the thermal equilirium with atomsphere at temperature `T` was found to be 11.08 atm. The moles fraction of `C_(3) H_(8)` in the mixture isA. 0.25B. 0.75C. 0.45D. 0.55 |
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Answer» Correct Answer - A `underset(a "mol")(C_(3) H_(8)) + underset(5a "mol")(5O_(2)) rarr underset(3a "mol")(3CO_(2)) + underset(-)(4H_(2) O(l))` `underset(b "mol")(C_(2) H_(4)) + underset(3b "mol")(3O_(2)) rarr underset(2b "mol")(2CO_(2)) + underset(-)(4H_(2) O (l)` Initially, `PV = nRT` `4.93 xx V = (a + b) RT` After combustion, pressure is due to the total moles of `CO_(2)` `11.8 xx V = (3a + 2b) RT` Divide equation (ii) by equation (i), we get `(11.08)/(4.93) = 2.25 = (3a + 2b)/(a + b)` `2.25 a + 2.25b = 3a + 2b` `0.25b = 0.75a` `b = 3a` `chi_(C_(2)H_(8))` or `chi_(a) = (a)/(a + b) = (a)/(a + 3a) = (1)/(4) = 0.25` |
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| 1023. |
Carbon and hydrogen combine to form three compounds A, B, and C. The percentage of hydrogen in these compounds are : `25,14.3 and 7.7` respectively. Show that the data illustrates the Law of Multiple Proportions. |
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Answer» Let us fix `1g` of hydrogen as the fixed weight in the three compounds In the first compound Weight of hydrogen `= 25.0 g` Weight of carbon `= 100 -25 = 75.0 g` `25.0 g` of hydrogen have combined with carbon `= 75.0 g` `1.0 g` of hydrogen has combined with carbon `= (75.0)/(25.0)g=3g` In the second compound Weight of hydrogen `=14.3 g` Weight of carbon `= 100 - 14.3 = 85.7 g` `14.3 g` of hydrogen have combined with carbon `= 100 - 14.3 = 85.7 g` `14.3 g` pf hydrogen have combined with carbon `= 85.7 g` `1.0 g` of hydrogen has combined with carbon `= (85.7)/(14.3)g=6g` In the third compound Weight of hydrogen `= 7.7 g` Weight of carbon `= 100-7.7 = 92.3 g` `7.7 g` of hydrogen have combined with carbon ` = 92.3 g` `1.0 g` of hydrogen has combined with carbon `= (92.3)/(7.7) g = 12 g` Ratio by weight of carbon which combine with `1g` of hydrogen in the three compounds is `3:6:12 or 1:2:4`. As the ratio is simple whole number in nature, the Law of Multiple Proportions is proved. |
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| 1024. |
A mixture of a mol of `C_(3) H_(8)` and `b` mol of `C_(2) H_(4)` was kept is a container of `V L` exerts a pressure of 4.93 atm at temperature `T`. Mixture was burnt in presence of `O_(2)` to convert `C_(3) H_(8)` and `C_(2) H_(4)` into `CO_(2)` in the container at the same temperature. The pressure of gases after the reaction and attaining the thermal equilirium with atomsphere at temperature `T` was found to be 11.08 atm. The moles of `O_(2)` needed for combustion at temperature `T` is equal toA. `14 a`B. `14 b`C. `15 a`D. `12 b` |
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Answer» Correct Answer - A moles of `C_(5) H_(8) +` moles of `C_(2) H_(4) = a + b` `= a + 3a = 4a` Mole of `O_(2) = 5a + 3b = 5a + 3 xx 3a = 14 a` |
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| 1025. |
Which of the statements are false?A. Physical quantity represented by volume is `dm^(3)`B. The length of pencil is `5 "cms"`.C. The work done by a system is 5 Joules.D. Air sometimes is considered as a hetrogeneous mixture due to the presence of dust particles which form a separate phase. |
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Answer» Correct Answer - B::C `implies 5 cm` (not cms) `implies 5` joules (not Joules) |
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| 1026. |
`1.80g` of a certain metal burnt in oxygen gave `3.0g` of its oxide `1.50g` of the same metal heated in steam gave `2.50g` of its oxide. Show that these illustrate the law of constant proportion . |
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Answer» In the first sample of the oxide, wt of metal `=1.80g, " wt of oxygen" = (3.0 - 1.80) g = 1.2g` `:. ("wt. of metal")/("wt.of oxygen") = (1.80g)/(1.2g) = 1.5` In the second sample of the oxide, wt of metal `=1.50g` wt of oxgen `= (2.50 - 1.50) g = 1g` `:. ("wt.of metal")/("wt of oxygen") = (1.50g)/(1g) = 1.5` Thus in both smaples of the oxide the proportions of the weights of the metal and oxgen are fixed. Hence, the results follows the law of constant proportion Note This law is not applicable in isotopes . |
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| 1027. |
A sample contains a mixtrure of `NaHCO_(3)` and `Na_(2)CO_(3)`. `HCl` is added to `15.0 g`.of the sample, yielding `11.0 g` of `NaCl`. What percent of the sample is `Na_(2)CO_(3)`? `[{:("Reaction are"),(Na_(2)CO_(3) + 2HCl rarr 2NaCl + CO_(2) + H_(2) O),(NaHCO_(2) + HCl rarr NaCL + CO_(2) + H_(2) O):}]` `Mw "of" NaCl = 58.5, Mw "of" NaHCO_(3) = 84, Mw "of" Na_(2)CO_(3) = 106 g mol^(-1)` |
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Answer» Let `x` of `Na_(2)CO_(3)`. Then, weight of `NaHCO_(3) = (15 - x) g` Moles of `NaCl` produced `= (11.0 g)/(58.5 g) = 0.188"mol"` The `NaCl` is produced by the reaction of `((x)/(106)) "mol"` of `Na_(2) CO_(3)` and `((15 - x))/(84) "mol of" NaHCO_(3)`. Each mol of `Na_(2)CO_(3)` produces 2 mol of `NaCl` `:. (2x)/(106) + (15 - x)/(84) = 0.188` Solve `x : = 13.5 g Na_(2) CO_(3)`, `NaHCO_(3) = (15 - 1.35) = 13.6 g` `% Na_(2)CO_(3) = (1.35)/(15) xx 100 = 9.0% Na_(2) CO_(3)` |
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| 1028. |
To prepare `100 g` of a 92% by weight solution of `NaOH` how many `g` of `H_(2) O`is needed? |
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Answer» Weight of `NaOH = ((92 g)/(100 g "solution")) (100 g "solution")` `= 92 g NaOH` Weight of `H_(2) O = 100 - 92 = 8.0 g H_(2) O` |
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| 1029. |
When 4.2 g `NaHCO_(3)` is added to a solution of `CH_(3)COOH` weighing 10.0 g, it is observed that 2.2 g `CO_(2)` is released into atmosphere . The residue is found is weigh 12.0 g . Show that these observations are in argreement with the law of conservation of mass. |
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Answer» `NaHCO_(3) + CH_(3)COOH to CH_(3)COONa + H_(2)O + CO_(2)` Initial mass =4.2 + 10=14.2 Final mass=12+2.2=14.2 Thus, during the course of reaction law of conservation of mass is obeyed |
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| 1030. |
Salt cake `(Na_(2)SO_(4))` is prepared as follows: `2NaCl + H_(2) SO_(4) rarr Na_(2)SO_(4) + 2HCl` How much salt cake could be produced from `100.0 g` of 90% pure saltin the above reaction?A. `109.8 g`B. `54.9 g`C. `36.6 g`D. `209.8 g` |
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Answer» Correct Answer - A Weight of `Na_(2)SO_(4) = (100 g NaCl)` `((90.0 g NaCl)/(100 g "mixture"))((1 "mol" NaCl)/(58.5 g NaCl))` `((1 "mol" Na_(2)SO_(4))/(2 "mol"NaCl))((142 g NaSO_(4))/("mol"Na_(2)SO_(4)))` `= (100 xx 90 xx 1 x 1 x 142)/(100 xx 58.5 xx 2)` `= 109.79 ~~ 109.8 g` |
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| 1031. |
Nitric acid is the most important oxyacid formed y nitrogen .It is one of the major industial chemicl and is widely used. Nitric acid is manufactured by the catalytic oxidation of ammonia in what is known as OSTWALD PROCESS which can be represented by the sequenve of reactions shown below: `4NH_(3)(g) +5O_(2)(g) underset("Catalyst")overset(Pt//Rh)rarr 4NO(g)+6H_(2)O(g) " "...(i)` `2NO(f) + O_(2)(g) overset(1120 K) rarr 2NO_(2)(g)" "...(ii)` `3NO_(2)(g)+H_(2)O(l) rarr 2HNO_(3)(aq)+NO(g)" "...(iii)` The aqueous nitric acid obtained by this method can be concentrated by distillation to ~ 68.5 % by weight . Further concentrated to 98% acid can be achieved by dehyration with concentrated sulphuric acid. If 170 kg of `NH_(3)` is heated in excess of oxygen then the volume fo `H_(2)O(l)` produced in 1st reaction at STP si , `(rho_(H_(2))O=1 gm//ml))`A. `340 xx 10^(3)L`B. 270LC. `227 xx 10^(3) L`D. 170L |
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Answer» Correct Answer - B |
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| 1032. |
Nitric acid is the most important oxyacid formed y nitrogen .It is one of the major industial chemicl and is widely used. Nitric acid is manufactured by the catalytic oxidation of ammonia in what is known as OSTWALD PROCESS which can be represented by the sequenve of reactions shown below: `4NH_(3)(g) +5O_(2)(g) underset("Catalyst")overset(Pt//Rh)rarr 4NO(g)+6H_(2)O(g) " "...(i)` `2NO(f) + O_(2)(g) overset(1120 K) rarr 2NO_(2)(g)" "...(ii)` `3NO_(2)(g)+H_(2)O(l) rarr 2HNO_(3)(aq)+NO(g)" "...(iii)` The aqueous nitric acid obtained by this method can be concentrated by distillation to ~ 68.5 % by weight . Further concentrated to 98% acid can be achieved by dehyration with concentrated sulphuric acid. If 180 litre of water completely reacts with `NO_(2)` produced to form nitric acid according ot the above reactions then the volume of air at STP containing 20% of `NH_(3)` is :(`rho_(H_(2)O)=1 gm//ml`)A. `1.56xx 10^(6) L`B. `6.81 xx 10^(4)L`C. `3.40 xx 10^(6)L`D. none of these |
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Answer» Correct Answer - C |
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| 1033. |
Nitric acid is the most important oxyacid formed y nitrogen .It is one of the major industial chemicl and is widely used. Nitric acid is manufactured by the catalytic oxidation of ammonia in what is known as OSTWALD PROCESS which can be represented by the sequenve of reactions shown below: `4NH_(3)(g) +5O_(2)(g) underset("Catalyst")overset(Pt//Rh)rarr 4NO(g)+6H_(2)O(g) " "...(i)` `2NO(f) + O_(2)(g) overset(1120 K) rarr 2NO_(2)(g)" "...(ii)` `3NO_(2)(g)+H_(2)O(l) rarr 2HNO_(3)(aq)+NO(g)" "...(iii)` The aqueous nitric acid obtained by this method can be concentrated by distillation to ~ 68.5 % by weight . Further concentrated to 98% acid can be achieved by dehyration with concentrated sulphuric acid. 85 kg of `NH_(3)(g)` was heated with 320 kg oxygen in the first step and `NHO_(3)` is prepared according to the above reactions . If the above reactions . If the final solution has volume 500 L ,then molarity of `HNO_(3)` is : [Assume NO formed finally is not reused]A. 2 MB. 8 MC. 3.33 MD. 6.66 M |
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Answer» Correct Answer - D |
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| 1034. |
If the masses of `Mn` of `O` are in the ratio of `55 : 16` in `MnO`, what is the ratio of `O` that combines with the same mass of `Mn` in `MnO_(2)` and `Mn_(2) O_(7)` ? |
| Answer» In `MnO_(2)`, 1 atom of `Mn` combines with 1 atom of `O (16)`. Therefore,in `MnO_(2)`, 1 atom of `Mn` will combine with 7 atoms or `7 xx 16 = 112` parts of `O` in `Mn_(2)O_(7)`. Hence, 1 atom of `Mn` will combine with `112//2 = 56` parts of oxygen. The ratio of `O` that combines with the same mass of `Mn` is `MnO_(2) : Mn_(2)O_(7) : : 32 : 56`, i.e., `4 : 7`. | |
| 1035. |
A trivalent metal has mass % of metal in its superoxide equal to 36%. Calcualte approximate specific heat capacity of metal.A. `(6.4/18) Cal//gm ^(@)C`B. `6.4/18 Cal//gm ^(@)C`C. `(18/6.4) Cal//gm ^(@)C`D. `(54/6.4) Cal//gm ^(@)C` |
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Answer» Correct Answer - B |
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| 1036. |
Calculate the molarity of water if its density is `1000 kg//m^(3)`. |
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Answer» Density of water `= ((1000kg))/(m^(3))=((100xx10^(3)g))/(("100 cm")^(3))=1 g cm^(-3)=1 g mL^(-1)` Mass of 1L(1000mL) of water `=(1000 mL)xx(1g mL^(-1))=100g` Molarity of water `= ("Mass of water/Molar mass")/("Volume of water")` `=((1000g))/((18"g mol"^(-1))xx"1 L")=55.56 "mol L"^(-1)=55.56M`. |
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| 1037. |
Equal masses of oxygen, hydrogen and methane are taken in a container in identical conditions. Find the ratio of their moles.A. `1:16:2`B. `1:16:1`C. `16:1:1`D. `16:2:1` |
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Answer» Correct Answer - 1 `w=mol_(O_(2))xx32` `w=mol e_(H_(2))xx2` `w=mol e_(CH_(4))xx16` `w=mol_(O_(2))xx32=mol e_(H_(2))xx2=mol e(CH_(4))xx16` `=1:16:2` |
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| 1038. |
Calculate the molarity of water if its density is `1000 kg m^(-3)` |
| Answer» Correct Answer - `55.5 "mol" L^(-1)` | |
| 1039. |
A `10 mL` sample of human urine was found to have `5 mg` of urea on analysis. Calculate the molarity of the given sample w.r.t. urea. (molecular mass of urea = 60) |
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Answer» `M = (W_(2) xx 1000)/(Mw_(2) xx V_(sol) ("in" mL))` `W_(2) = 5 g = 5xx10^(-3) g` `:. M = (5 xx 10^(-3) xx 10^(3))/(60 xx 10) = 0.008` |
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| 1040. |
`5L` of an alkane requires `25L` of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure, the alkane is :A. ButaneB. IsobutaneC. PropaneD. Ethane |
| Answer» Correct Answer - C | |
| 1041. |
How many moles of electrons weigh `1 kg`?A. `6.023xx10^(23)`B. `(1)/(9.108)xx10^(31)`C. `(6.023)/(9.108)xx10^(54)`D. `(1)/(9.108xx6.023)xx10^(8)` |
| Answer» Correct Answer - D | |
| 1042. |
At `100^(@)C` and 1 atm, if the density of the liquid water is `1.0 g cm^(-3)` and that of water vapour is `0.0006 g cm^(-3)` , then the volume occupied by water molecules in `1 L` steam at this temperature isA. `6 cm^(3)`B. `60 cm^(3)`C. `0.6 cm^(3)`D. `0.06 cm^(3)` |
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Answer» Correct Answer - D Mass of water in `1 L` steam `= 1000 xx 0.00006 g` `= 0.06 g` Volume of `0.06 g` water `= (0.06)/(1 g cm^(-3)) = 0.06 cm^(3)` |
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| 1043. |
At `100^(@)C` and 1 atm, if the density of the liquid water is `1.0 g cm^(-3)` and that of water vapour is `0.0006 g cm^(-3)` , then the volume occupied by water molecules in `1 L` steam at this temperature isA. `6cm^(3)`B. `60cm^(3)`C. `0.6cm^(3)`D. `0.06cm^(3)` |
| Answer» Correct Answer - C | |
| 1044. |
Calculate the density of `NH_(3)` at `30^(@)C` and 5 atm pressure.A. `3.42 g L^(-1)`B. `2.42 g L^(-1)`C. `1.71 g L^(-1)`D. `3.84 g L^(-1)` |
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Answer» Correct Answer - A `P = (dRT)/(M)` `d = (PM)/(Rt) = (5 xx 17)/(0.0832 xx 300) = 3.42 g L^(-1)` |
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| 1045. |
Which one of the following parts of gases contains the same number of molecules?A. `16g " of "O_(2)and14 g " of "N_(2)`B. `8g " of "O_(2)and22 g " of " CO_(2)`C. `28 g " of " N_(2)and 22 g " of " CO_(2)`D. `32g " of " O_(2)and 32g " of " N_(2)` |
| Answer» Correct Answer - 1 | |
| 1046. |
Calculate the density of `NH_(3)` at `30^(@)C` and 5 atm pressure. |
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Answer» Correct Answer - A::C::D Density of a gas `= (PM)/(RT) = (5 xx 17)/(0.0821 xx 303) = 3.417 g L^(-1)` |
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| 1047. |
The number of oxygen atoms in 4.4 g of `CO_(2)` is approximatelyA. `1.2xx10^(23)`B. `6xx10^(22)`C. `6xx10^(23)`D. `12xx10^(23)` |
| Answer» Correct Answer - 1 | |
| 1048. |
Which of the following solution contains same molar concentration?A. `166 g. KI//L` solution.B. `33.0 g (NH_(4))_(2) SO_(4)` in `200 mL` solutionC. `25.0 g CuSO_(4).5H_(2) O` in `100 mL` solutionD. `27.0 mg Al^(3+)` per `mL` solution. |
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Answer» Correct Answer - A::C::D `[{:(Mw "of" KI (NH_(4))_(2) SO_(4) CuSO_(4) 5H_(2) "and"),(Al^(2+) "respectively are" 166 132 250 "and"),(27 g mol^(-1)):}]` a. `M = (166 xx 1000)/(166 xx 1000) = 1.0 M` b. `M = (33 xx 1000)/(132 xx 200) = 1.25 M` c. `M = (25 xx 1000)/(250 xx 100) = 1.0 M` d. `M = (27 xx 10^(-3) xx 1000)/(27 xx 1) = 1.0 M` |
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| 1049. |
The weight of `1xx10^(22)` molecules of `CuSO_(4).5H_(2)O` isA. 14.59 gB. 415 .9 gC. 4. 159 gD. None of these |
| Answer» Correct Answer - 3 | |
| 1050. |
Assertion `:` A mixture of plant pigments can be separated by chromatography. Reason `:` Chromatography is used for the separation of colourd substances into individual components.A. CalorimetryB. ChromatographyC. CalorimetryD. Gravimetry |
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Answer» Correct Answer - B |
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