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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.
| 251. |
A mixture of `CaCl_(2)` and NaCl weighing 4.44 is treated with sodium carbonate solution to precipitate all the `Ca^(2+)` ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get 0.56 g of `CaO`. The percentage of NaCl in the mixture of (atomic mass of Ca=40) isA. `75`B. `30.6`C. `25`D. `69.4` |
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Answer» Correct Answer - A `NaCl+Na_(2)CO_(3)to` No reaction `CaCl_(2)+Na_(2)CO_(3)toCaCO_(3)+2NaCl` `underset(=111g)((40+2xx35.5))underset(=106g)((23xx2+12+16xx3))underset(=100g)((40+12+3xx16))` Let the weight of `CaCl_(2)` in the mixture be x g `:.` weight of `NaCl` in the mixture `=(4.44-x)g` `CaCO_(3)overset(Delta)(to)CaO+CO_(2)` `40+16` `=56g` Now 111 g `CaCl_(2)` produce `CaCO_(3)=100g` `:.` x g of `CaCO_(2)` produce CaO = 56 g `(100)/(111)xCaCO_(3)` produce CaO `= (56)/(100)xx(100)/(111)xg` `=(56)/(111)xg=0.50g` or x = 1.11 g `:.` NaCl in the mixture = 4.44-1.11 = 3.33 g Percentage of NaCl in the mixture `=(3.33)/(4.44)xx100=75%` |
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| 252. |
If `0.50 mol` of barium chloride `(BaCl_(2))` is mixed with `0.20` mol of sodium phosphate `(Na_(3) PO_(4))` the maximum number of moles of barium phosphate `[Ba_(3)(PO_(4))_(2)]` formed isA. `0.25`B. `0.10`C. `0.40`D. `0.50` |
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Answer» Correct Answer - B `underset(3 mol)(3BaCl_(2)) + underset(2 mol) (2Na_(3) PO_(4)) rarr underset(1 mol) (Ba_(3)(PO_(4))_(2)) + underset(6 mol) (6NaCl)` `0.50 mol BaCl_(2) xx (1 mol Ba_(3) (PO_(4))_(2))/(3 mol BaCl_(2)) = 0.17 mol Ba_(3) (PO_(4))_(2)` `0.20 mol Na_(3) PO_(4) xx (1 mol Ba_(3) (PO_(4))_(2))/(2.0 mol Na_(3) PO_(4)) = 0.1 mol Ba_(3) (PO_(4))_(2)` Therefore, `Na_(3) PO_(4)` is the limiting rectant and the maximum moles of `Ba_(3) (PO_(4))_(2)` formed is `0.1` mol. |
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| 253. |
Consider the following reactioin, 2Al(s) + 6HCl(aq) `to` `2Al^(3+)(aq) + 6Cl^(-)(aq) + 3H_(2)(g)` Which of the following statements is incorrect?A. 6 L HCl (aq) is consumed for every `3LH_(2)`(g) producedB. 33.6 L `H_(2)`(g) is produced regardless of temperature and pressure for every mole of Al that reactsC. 67.2 L `H_(2)` L `H_(2)`(g) at STP, is produced for every mole of AI that reactsD. 11.2 L `H_(2)` (g) at STP, is produced for every mole of HCl (aq) consumed |
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Answer» 2 moles of Al produce `=3xx22.4L`of `H_(2)` gas `therefore`1 mole of Al will produce `=(3xx22.4)/(2)=33.62L` of `H_(2)` gas |
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| 254. |
The mass of `2.24xx10^(-3)m^(3)` of a gas is 4.4 g at 273.15 K and 101.325 Kpa pressure. The gas may beA. `NO`B. `NO_(2)`C. `C_(3)H_(8)`D. `NH_(3)` |
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Answer» Correct Answer - C `2.24xx10^(-3)m^(3)` of a gas = 2.24L at STP `:. 2.24L` of a gas at STP weigh = 4.4 `:.` 22.4 L of a gas at STP weigh =44g now 22.4 L of a gas at STP at = molar mass of the gas `:.` molar mass of the gas = 44 g molar mass of `C_(3)H_(8)=36+8=44g` |
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| 255. |
Assertion: `HClO_(4)` is a stronger acid than `HClO_(3)`. Reason: Oxidation state of `Cl` in `HClO_(4)` is `+VII` and in `HClO_(3)+V`.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - B Both assertion and reason are true but reason is not the correct explanation of assertion. Greater the number of negative atoms present in the oxy-acid make the acid stronger. In general, the strength of acids that have general formula `(HO)_(m)ZO_(n)` can be related to the value of n. As the value of n increases, acidic character also increases. The negative atoms draw electrons away from the Z-atom and make it more positive. The Z-atom, therefore, becomes more effective in with drawing electron density away from the oxygen atom that bonded to hydrogen. in turn, the electrons of `H-O` bond are drawn more strongly away from the H-atom. The net effect makes it easier from the proton release and increases the acid strength. |
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| 256. |
There are two isotopes of an element with atomic mass z. Heavier on has atomic mass z+2 and lighter one has z-1, then abundance of lighter one isA. `66.6%`B. `96.7%`C. `6.67%`D. `33.3%` |
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Answer» Correct Answer - A Let percent abundance of lighter isotope = x `:. (x(z-1)+(100-x)(z+2))/(100)=z` `xz-z+100z+200-xz-2x=100z` `-3x = -200` `x = (200)/(6) = 66.6` `:.` percent abundance of lighter isotopes = x |
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| 257. |
Boron has two isotopes, B-10 and B-11. The average atomic mass of boron is found to be 10.80u. Calculate the percentage of abundance of these isotopes. |
| Answer» Correct Answer - B-10 = 20%, B-11 =80% | |
| 258. |
Assertion: Matter can neither be created nor be destroyed. Reason : This is law of definite proportions.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Law of conservation of mass state that matter can neither be created nor destroyed. | |
| 259. |
Write two challenges to chemists of future. |
| Answer» Management of green house gases, Bio-chemical processes. | |
| 260. |
Classify the following as pure substances or mixtures, give reasons.A. PetrolB. Tap waterC. Distilled waterD. Oxygen |
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Answer» Pure substance : Distilled water and oxygen have fixed composition. Mixture : Petrol and tap water do not have fixed composition. |
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| 261. |
A tumbler contains 3 L of water Calculate the volume of the water in `m^(3)`. |
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Answer» We know that, 1 m-100 cm and `1 L - 10^(3)cm^(3)` To get `m^(3), 1 m^(3)=(100 cm)^(3)=10^(6)cm^(3)` As, `1 L = 10^(3) cm^(3)` `therefore 1 m^(3)=(10^(6)cm^(3))/(10^(3) cm^(3))=10^(3) L` Hence, `1 L=10^(-3) m^(3)` `therefore 3 L` of water `= 3xx10^(-3)m^(3)` OR Since, `1 L = 1000 cm^(3)` and 1 m = 100 cm which gives `(1 m)/(100 cm) = 1 = (100 cm)/(1 m)` To get `m^(3)` from the above unit factors, the first unit factor is taken and it is cubed. `((1 m)/(100 cm))^(3) rArr (1 m^(3))/(10^(6)cm^(3))=(1)^(3)=1` Now, `3 L = 3xx1000 cm^(3)` The above is multiplied by the unit factor. `3xx1000 cm^(3)xx(1m^(3))/(10^(6)cm^(3))=(3 m^(3))/(10^(3))=3xx10^(-3)m^(3)`. |
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| 262. |
Express the result of the following calculations to the appropriate number of significant digits. `6.66xx3.33+216.67` |
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Answer» `6.66xx3.33+216.67` `{:(" 22.18"),(ul("+216.67")),(ul(" 23885")):}` After rounding off the result is 238. |
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| 263. |
Express the following calculations to the proper number of significant figures. `((2.34xx10^(-8))(0.5))/(6.4)` |
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Answer» In this calculation 0.5 has minimum number of significant digits. It has only one significant digit. The result of this calculation is, therefore , to be rounded off to one significant digit. `((2.34xx10^(-8))xx(0.5))/(6.4)` `=(1.17xx10^(-8))/(6.4)=0.1828125xx10^(-8)` Here 8, after 1 is greater than 5, so the result after rounding off is `0.2xx10^(-8)` . |
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| 264. |
Express the following numbers to four significant figures. (1) `1.81234xx10^(3)` (2) 0.008837 |
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Answer» (1) The fifth digit is less than 5, therefore result is expressed as `1.812xx10^(3)` (2) The four digits to be retained are 8.3 and 7 therefore, the number is expressed as `8.837xx10^(-3)`. Note : The exponential terms does not add to significant figures. |
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| 265. |
Express the following number to two significant figures (i) 5.602792 (ii) 3.3402800 |
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Answer» Correct Answer - (i) 5.6 (ii) 3.3 |
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| 266. |
After rounding of 5.235 and 5.225 to three significant figures , we will have answers respectively asA. `5.23, 5.22`B. `5.24, 5.123`C. `5.23, 5.23`D. `5.24, 5.22` |
| Answer» Correct Answer - D | |
| 267. |
What will be the length of a metre stick of 5.602 m when 2.8 m portion is cut from it ? Express the result to the appropriate significant figures. |
| Answer» Correct Answer - 2.8 m | |
| 268. |
Given the number `786*, 0*786` and `0*0786`. The number of significant figures for the three numbers isA. 3, 4 and 5 respectivelyB. 3, 3 and 3 respectivelyC. 3, 3 and 4 respectivelyD. 3, 4 and 4 respectively |
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Answer» Correct Answer - B Zeros to the left of the first non-zero digit in a number are not significant |
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| 269. |
On dividing 0*25 by 22*1176, the actual answer is 0*011303. The correctly reported answer will beA. `0*011`B. `0*01`C. `0*0113`D. `0*013` |
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Answer» Correct Answer - A `0*25÷22*1176=0*011303`. As the least precise term involved has two significant figures (viz., `0*25`) therefore reported sum is 599 |
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| 270. |
The actual product of 5.786 and 4.2 is 24.3012 . The correctly reported answer will beA. 24B. `24.3`C. `24.30`D. `24.301` |
| Answer» Correct Answer - A | |
| 271. |
The correctly reported answer of the addition of `294*406, 280*208` and 24 will beA. `598*61`B. `599*`C. `598*6`D. `598*614` |
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Answer» Correct Answer - B `294*406 + 280*208 + 24 = 598*614`. As 24 has no decimal place i.e., it is an exact number. After rounding off, reported sum is 599 |
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| 272. |
On dividing 0.86 by 16.564, the actual answer is 0.0519198. The correctly reported answer will beA. `0.05`B. `0.051`C. `0.0519`D. `0.05191` |
| Answer» Correct Answer - B | |
| 273. |
Stoichiometric ratio of sodium dihydrogen orthophosphate and sodium hydrogen orthophosphate required for synthesis of `Na_(5) P_(3) O_(10)` isA. `1.5 :3`B. `3 : 1.5`C. `1 : 1`D. `2 : 3` |
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Answer» Correct Answer - A `underset(underset("orthophosphate ")("Sodium dihydrogen"))(3NaH_(2) PO _(4)) + underset(underset("orthophosphate ")("Sodium hydrogen "))(6Na_(2) HPO_(4))rarr 3Na_(5)P_(3) O_(10) + 6H_(2) O ` Hence, stoichiometric ratio = `1.5 :3` |
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| 274. |
Which of the following is Loschmidt numberA. `6xx10^(23)`B. `2.69xx10^(19)`C. `3xx10^(23)`D. None of these |
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Answer» Correct Answer - B The no. of molecules present in 1 ml of get at STP is known as Loschmidt number. 22400 ml of gas has total no. of molecules `=6.023xx10^(23)` 1 ml of gas has total no. of molecules `=(6.023xx10^(23))/22400` `=2.69xx10^(19)`. |
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| 275. |
The normally of `0.3 M` phosphorus acid `(H_(3) PO_(3))` isA. `0.6 N`B. `0.3 N`C. `0.9 N`D. `0.1 N` |
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Answer» Correct Answer - A Structure of `H_(3) PO_(3) (H-overset(O)overset(||)underset(OH)underset(|)(P)-OH)` shows that only two `H` atoms are bonded to the hightly electronegative `O` atom. Thus, it is a dibasic (diprotic) acid,n i.e., it can furnish only two `H` atoms per molecule as `H^(+)` ions in aqueous solution. Its `n` factor is thus, 2 and `N = 2M = 2(0.3) = 0.6` |
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| 276. |
Which one of the following will have the largest number of atoms? (i) 1 g Au (s) (ii) 1 g Na (s)(iii) 1 g Li (s) (iv) 1 g of Cl2(g)A. 1 g Au (s)B. 1 g Na (s)C. 1 g Li (s)D. 1 g of `Cl_(2)(g)` |
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Answer» Correct Answer - C Number of moles `=("mass")/("atomic masss")="x moles" =xN_(0)" atom"` Thus, atoms `prop 1/("atomic mass")` Least atomic mass is that of lithium. Thus, Li have largest number of atoms. |
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| 277. |
If `m_(1)` and `m_(2)` are masses of two reactants in any reaction, having their gram equivalent masses `E_(1)` and `E_(2)` respectively, which of the following equatios represents the law of equivence correctly?A. `m_(1)/m_(2)=E_(2)/E_(1)`B. `E_(1)E_(2)=m_(1)m_(2)`C. `m_(1)E_(2)=E_(1)m_(2)`D. `(m_(1)+m_(2))=(E_(1)+E_(2))` |
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Answer» Correct Answer - C The correct formulae that expresses the law of equivalent correctly is `m_(1)E_(2)=E_(1)m_(2)` |
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| 278. |
For the redox reaction `MnO_(4)^(-)+C_(2)O_(4)^(-2)+H^(+) rarr Mn^(2+)+CO_(2)+H_(2)O` the correct coefficients of the reactants for the balanced reaction areA. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(2,5,16):}`B. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(16,5,2):}`C. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(5,16,2):}`D. `{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(2,16,5):}` |
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Answer» Correct Answer - A `MnO_(4)^(-) + 8H^(+) + 5e^(-) rarr Mn^(2+) + 4H_(2) O xx 2` `(c_(2)O_(4)^(2-) rarr 2 CO_(2) + 2e^(-) xx 5)/(2MnO_(4)^(-) + 5C_(2)O_(4)^(2-) + 16H^(+) rarr 2Mn^(2+) + 10CO_(2) + 8H_(2)O)` Thus the coefficient of `MnO_(4)^(-), C_(2)O_(4)^(2-) and H^(+)` in the abovve balanced equation respectively are 2, 5, 16 |
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| 279. |
Which of the following is correct? (i) `(5.28xx0.156xx3)/(0.0428) = 55.7` (ii) `(5.28xx0.156xx3)/(0.0421) = 56.7` (ii) `(42.967xx0.02435)/(034xx4) = 0.77`A. (i),(ii)B. (i),(ii),(iii)C. (i),(iii)D. (ii),(iii) |
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Answer» Correct Answer - B In all the cases, the reported quotient has the minimum number of significant figures. |
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| 280. |
Assertion: Volume of a gas is inversely proportional to the number of moles of a gas. Reason: The ratio by volume of gaseous reactants and products is in agreement with their molar ratio.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - D We known that from the reaction `H_(2)+Cl_(2) rarr 2HCl` that the ratio of the volume of gaseous reactants and products is in agreement with their molar ratio. The ratio of `H_(2) : Cl_(2) : HCl` volumes id `1 : 1 : 2` which is the same as their number of moles. Therefore, the assertion is false but reasonis true. |
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| 281. |
Which statement is NOT true for the given reaction `Fe^(3+)+e^(-) to Fe^(2+)` ?A. `Fe^(3+)` being reducedB. Oxidation state of Fe has changedC. `Fe^(3+)` could be referred to an oxidising agent in this reactionD. Both `Fe^(3+)` and `Fe^(2+)` are called acid radicals |
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Answer» Correct Answer - D Both `Fe^(3+)` and `Fe^(2+)` are basic radicals. |
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| 282. |
When `HNO_(3)` is converted into `NH_(3)`, the equivalent weight of `HNO_(3)` will be:A. (a)`M//2`B. (b)`M//1`C. (c )`M//6`D. (d)`M//8` |
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Answer» Correct Answer - D `Hoverset(+5)(NO_(3)) rarr overset(-3)(NH_(3)) :. V.f. of HNO_(3)=8` `Eq. wt. =M//8`. |
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| 283. |
What is the atomic weight of an element X for which a sample containing `1.58xx10^(22)` atoms weigh 1.05 g?A. 28 gB. 20 gC. 40 gD. 23 g |
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Answer» Correct Answer - C We know that one mole of an element `= 6. 02 xx 10^(23)` atoms = atomic weight of the element Now, `1.58xx 10^(22)` atoms weigh `.05` g `therefore 6.02 xx 10^(23)` atoms weigh `= ((1.05xx6.02xx10^(23)))/(1.58 xx 10^(22)) = 40 ` g Atomic weight of `X=40` g |
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| 284. |
In the reaction `Pb(s)+Cu^(2+) (aq) rarr Pb^(2+) (aq)+Cu(s)` which is reducing agentA. `Pb^(2+) (aq)`B. `Cu^(2+) (aq)`C. `Pb(s)`D. `Cu(s)` |
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Answer» Correct Answer - C `Pb_((s))` releases `2e^(-)` to form `Pb_((aq))^(2+)` and its O.N. increases by 2, hence it is a reducing agent. |
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| 285. |
The percentage of nitrogen in urea is about:A. 28B. 18C. 85D. 46 |
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Answer» Correct Answer - D The formula of urea = `NH_(2) CONH_(2)` Molar mass of urea `=14 + (2 xx 1) + 12 + 16 + 14 + (2 xx 1)` `=60 "g mol"^(-1)` Nitrogen = 28 g `"mol"^(-1)` `therefore %` of N_(2) in urea `= 28/60 xx 10 = 46.6%cong 46%` Hence, the percentage of `N_(2)` in urea `=46.6%cong 46%` |
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| 286. |
A sample was weighted using two different balances. The results were (i) `3.929 g` and (ii) `4.0 g`. How would the weight of the sample be reported?A. (a)` 3.929 g`B. (b)` 3 g`C. (c )` 3.9 g`D. (d)`3.93 g` |
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Answer» Correct Answer - D Round off the digit at 2nd position of decimal 3.929=3.93. |
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| 287. |
The diameter of the nucleus of an atom is measured in fermi (femto)meter (fm) which is equal toA. `10^(-18) m`B. `10^(21) m`C. `10^(-15) m`D. `10^(-24)m` |
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Answer» Correct Answer - C `10^(-18)` (atto), `10^(-21)` (zepto), `10^(24)` (yocto). |
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| 288. |
Which of the following has an ambiguous number of significant figures?A. `7.03`B. `500`C. `0.05`D. `705.7` |
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Answer» Correct Answer - B Zeros at the end of a number that does not contain a decimal point or may not be significant. |
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| 289. |
Number of molecules in 1 L of water is close toA. `18/(22.4 )xx10^(23)`B. `55.5 xx 6 . 023 xx10^(23)`C. `(6.023)/23.4 xx 10^(23)`D. `18xx 6. 023 xx 10^(23)` |
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Answer» Correct Answer - B 1 mole of water means 18 g of water which means `6.023xx 10^(23)` molecules. 1 L of water means = 1000 g 1000 g of water `= 1000/18 ` mol As 1mol of water `= 6. 02 xx10^(23)` molecules `therefore 1000/18` mol of water ` = 1000/18 xx 6.023xx 10^(23)` ` = 55.5 xx 6.023 xx 10^(23)` |
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| 290. |
A 400 mg iron capsule contains 100 mg of ferrous fumarate, `(CHCOO)_(2)`Fe. The percentage of iron present in it is approximatelyA. `8*2`B. `25%`C. `16%`D. unpredictable |
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Answer» Correct Answer - A Mol. Mass of `Fe(CHCOO)_(2) = 170` Fe in 100 mg of `Fe(CHCOO)_(2) = (56xx100)/(170)` = 32.9 mg Total Fe in 400 mg of capsule = 32.9 mg `:.` percentage of Fe in capsule `= (32.9xx100)/(400) = 8.2` |
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| 291. |
The volume of water to be added to `100 cm^(3)` of 0.5 `NH_(2)SO_(4)` to get decinormal concentration isA. `400 cm^(3)`B. `500 cm^(3)`C. `450 cm^(3)`D. `100 cm^(3)` |
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Answer» Correct Answer - A `100xx0.5 = V_(2)xx(1)/(10)` or `V_(2) = 10xx100xx0.5=500cm^(3)` |
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| 292. |
The reaction of calcium with water is represented by the equation `Ca+2H_(2)OtoCa(OH)_(2)+H_(2)` What volume of `H_(2)`, at STP would be liberated when 8g of calcium completely reacts with waterA. `4480 cm^(3)`B. `2240 cm^(3)`C. `1120 cm^(3)`D. `0.4 cm^(3)` |
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Answer» Correct Answer - A `underset(40)(Ca)+2H_(2)OtoCa(OH)_(2)+underset(22400cm^(3))(H_(2))` 8f of calcium will produce `= (22400xx8)/(40)` `=4480cm^(3)` |
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| 293. |
If one mole of rupees is distributed equally amongst all the poputlation of earth (6 billion), each person will get rupees (approximately)A. `1000000`B. `10000000`C. `10^(20)`D. `10^(14)` |
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Answer» Correct Answer - D Each person will get `=(6.02xx10^(23))/(6xx10^(9))` `=1.003xx10^(14)=10^(14)`(approx.) |
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| 294. |
Determine the number fo significant figures in the following measured quanttities: (i) `478 cm`, (ii) `7.01 g` (iiI) `0.852 m`, (iv) `0.034 kg` (v) `1.410xx10^(22)` atoms, (vi) `8000 mL`. Strategy: Use the simple rules mentioned above to count the number of significant figures. |
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Answer» (i) Three, using 1 (ii) Three, using rule(2) (ii) (iii) Three, using rule2 (iv) (iv) Two, using rule 2(i) (v) Four, using rule 1, (vi) This is an ambiguous case as the number of significant figures may be one `(8xx10^(3))`. two `(8.0xx10^(3))`, three `(8.00xx10^(3))`, or four `(8.000xx10^(3))`. |
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| 295. |
1.12 ml of a gas is produced at STP by the action of 4.12 mg of alcohole, with methyl magnesium iodide. The molecular mass of alcohol isA. `16.0`B. `41.2`C. `82.4`D. `156.0` |
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Answer» Correct Answer - C `underset(1mol)(ROH)+CH_(3)MgBrtounderset("at S.P.T.")underset(22400mL)(CH_(4)) uarr+BrMgOR` 22400 mL of `CH_(4)` at S.T.P. = 1 mol of ROH 1.12 mL of `CH_(4)` at S.T.P. `=(1molxx1.12mL)/(22400mL)` `=5xx10^(-5)`mol `5xx10^(-6)`mol of ROH `=4.12 mg` `=4.12xx10^(-3)g` 1 mol of ROH `=(4.12xx10^(-3))/(5xx10^(-5))` `=0.824xx10^(2)=82.4g` `:.` Mol. mass of ROH = 82.4 |
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| 296. |
Which of the following is related to chemical property?A. Dissolution of zinc in hydrochloric acidB. Dissolution of sugar in waterC. Dissolution of sulphur in carbon disulphideD. Dissolution of benzence in ethyl alcohol |
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Answer» Correct Answer - A It is chemical reaction: `Zn(s) + 2HCl(aq.) rarr ZnCl_(2) (aq.) + H_(2) (g)` |
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| 297. |
In which of the following pairs of compounds the ratio of `C, H` and `O` is sameA. (a)Acetic acid and methyl alcoholB. (b)Glucose and acetic acidC. (c )Fructose and sucroseD. (d)All of these |
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Answer» Correct Answer - B Glucose-`C_(6)H_(12)O_(6)` Ratio of `C`, `H` and `O=1:2:1` In acetic acid `CH_(3)-underset(O)underset(||)(C)-O-H` Ratio of `C`, `H` and O `1:2:1`. |
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| 298. |
Which of the following is not related to physical property?A. Melting of iceB. Flow fo current through copperC. Burning of magnesium in the oxygen of the airD. Boiling of water |
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Answer» Correct Answer - C Magnesium oxide, a while solid is formed. `2 Mg(s) + O_(2)(g) rarr 2 MgO(s)` |
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| 299. |
A mixture of salt and water can be separated byA. hand pickingB. distillationC. filrationD. crystallization |
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Answer» Correct Answer - B Evaporation and condensation of water leaves the solid salt behind. |
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| 300. |
On reduction with hydrogen, `3.6 g` of an oxide of matel left `3.2 g` of metal. If the vapour density of metal is `32`, the simplest formula of the oxide would beA. (a)`MO`B. (b)`M_(2)O_(3)`C. (c )`M_(2)O`D. (d)`M_(2)O_(5)` |
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Answer» Correct Answer - C As we know that Equivalent weight`=("weight of metal")/("weight of oxygen")xx8` `=32/0.4xx8=64` Vapour density`=(mol. Wt)/2` Mol. Wt.`=2xxV.D=2xx32=54` As we know that `n=("mol. wt")/("eq. wt")=64/64=1` Suppose, the formula of metal oxide be. `M_(2)O_(n)` Hence the formula of metal oxide`=M_(2)O`. |
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