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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.
| 801. |
What is correct for 10 g of `CaCO_(3)`?A. It contains 1 g atom of carbonB. It contains `0*3` g atoms of oxygenC. It contains 12 g of calciumD. It refers to `0*1` g-equivalent of `CaCO_(3)` |
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Answer» Correct Answer - B 10 g of `CaCO_(3) = (10)/(100)` `= 0*1 g` mol of `CaCO_(3)` `=0*1xx3` g-atom of oxygen `=4g Ca = 0*2 g eq.` |
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| 802. |
Number of atoms in 12 g `._(12)^(24)Mg` is equal toA. oxygen atoms in 11 g `CO_(2)`B. hydrogen atoms in 4 g `CH_(4)`C. nitrogen atoms in 46 g `N_(2) O_(4)`D. sulphur atoms in 79 g `Na_(2)S_(2) O_(3)` |
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Answer» Correct Answer - A Number of atoms = number of molecues `xx` specific atoms in molecules In` {:(24),(12):}` Mg number of atom `= =12/24 xx 6.022 xx 10^(23)xx1` `=3.011xx 10^(23)` atoms Number of `O_(2)` atoms in 11 `CO_(2)` `= 11/44 xx 6.022 xx 10^(23)xx2` `=3.011 xx 10^(23)` atoms |
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| 803. |
A signature written with carbon pencil weighs 1 mg. What is the number of carbon atoms present in the signature?A. `6.02xx10^(20)`B. `0.502xx10^(20)`C. `5.02xx10^(23)`D. `5.02xx10^(20)` |
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Answer» Correct Answer - B 12 g of C-12 contain `6.02xx10^(23)` atoms `10^(-3)` (1mg) of C-12 contain atoms `= (6.02xx10^(23))/(12)xx10^(-3) = 0.502xx10^(20)` |
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| 804. |
What is the total number of atoms present in 25.0 mg of camphor, `C_(10)H_(16)O`?A. `9.89xx10^(19)`B. `6.02xx10^(20)`C. `9.89xx10^(20)`D. `2.67xx10^(21)` |
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Answer» Correct Answer - D Molar mass of `C_(10)H_(16)O = 12+16+16 = 152 g mol^(-1)` `25.0 mg = 25.0 mgxx((1)/(10^(3)mg))xx((1mol)/(152g))xx(27xx6.02xx10^(23)atoms)/(1mol)` `=(25xx27xx6.02)/(152)xx10^(20)` atom `= 26.73xx10^(20)` atom `= 2.67xx10^(21)` atoms |
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| 805. |
Number of atoms in 12 g `._(12)^(24)Mg` is equal toA. oxygen atoms in 11 g `CO_(2)`B. hydrogen atoms in 4g `CH_(4)`C. nitrogen atom in 46 g `N_(2)O_(4)`D. sulphur atom in 79 g `Na_(2)S_(2)O_(3)` |
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Answer» Correct Answer - A 12 g of `._(12)^(24)Mg`(0.5 mol) contains atoms `= 3.01xx10^(23)` (A) 11 g of `CO_(2)`(0.25 mol) contains oxygen atom `= 0.50 mol` `=3.01xx10^(23)` atoms (B) 4 g of `CH_(4)`(0.25 mol) contain hydrogen atoms `= 1 mol = 6.02xx10^(23)`atoms (C) 46 g of `N_(2)O_(4)`(0.5 mol) contains sulphur atoms `=2 mol = 2xx6.02xx10^(23)` atoms (D) 79 g of `Na_(2)S_(2)O_(3)` (0.5 mol) contains sulphur atoms = 1 mol = `6.02xx10^(23)` atom (Molar mass of `CO_(2), CH_(4), N_(2)O_(4)` and `N_(2)S_(2)O_(3)` are 44, 16, 92 and `158 g mol^(-1)` |
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| 806. |
A sample of `NaCO_(3)` cotains `6*02xx10^(23) Na^(+)` ions. The mass of the sample is ( at. Mass Na=23, C=12, O=16)A. 53 gB. 106 gC. 165 gD. 212 g |
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Answer» Correct Answer - A No. of `Na^(+)` ions = `6*02xx10^(23)` `:.` No. of formula units of `Na_(2)CO_(3)` `=(6*02xx10^(23))/(2) = 3*01xx10^(23)` Molar mass of `Na_(2)CO_(3)` `=2xx23+12+16xx3` `= 46+12+48=106g mol^(-1)` `6*02xx10^(23)` formula units `Na_(2)CO_(3)` has mass `=106 g` `:. 3*01xx10^(23)` formula units of `NaCO_(3)` has mass `(106xx3*01xx10^(23))/(6*02xx10^(23))=52 g` |
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| 807. |
The number of significant figures in `6.02xx10^(23)` isA. (a)` 23`B. (b)` 3`C. (c )`4 `D. (d)` 26` |
| Answer» Correct Answer - B | |
| 808. |
Compute the value of x : x = 9.4g of phenol `(C_(6)H_(5)OH) +6.02xx10^(23)` molecules of phenol - 0.2 mole of phenol Here x is:A. 0.9 molB. 9.2 gC. 0.1 molD. `6.02xx10^(23)` molecules |
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Answer» Correct Answer - A Molar mass of phenol, `C_(6)H_(5)OH = 72+5+16+1 = 94 g mol^(-1)` 9.4 g of phenol = 0.1 mol `6.02xx10^(23)` molecules of phenol = 1.0 mol `:.` x = 0.1 mol + 1.0 mol - 0.2 mol = 0.9 mol |
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| 809. |
`KMnO_(4)` react with oxalic acid according to the equation, `2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+) rarr 2Mn^(2+)+ 10 CO_(2)+8H_(2)O`, here `20 ml` of `0.1 M KMnO_(4)` is equivalemt toA. (a)`20 "ml of" 0.5 M H_(2)C_(2)O_(4)`B. (b)`50 "ml of" 0.1 M H_(2)C_(2)O_(4)`C. (c )`50 " ml of" 0.5 M H_(2)C_(2)O_(4)`D. (d)`20 "ml of" 0.1 M H_(2)C_(2)O_(4)` |
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Answer» Correct Answer - B `KMnO_(4)` Oxalic acid `(M_(1)V_(1))/n_(1)=(M_(2)V_(2))/n_(2), (20xx0.1)/2=(M_(2)V_(2))/5, M_(2)V_(2)=5`. |
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| 810. |
Arrange the following in order of increasing mass (i) `3.0115xx10^(23)` molecules of white phosphorus (ii) 10 mole of `H_(2)` gas (iii) 1 g molecule of anhydrous `Na_(2)CO_(3)` (iv) `33.6 L` of `CO_(2)` gas at STPA. (a)`iilt iltIvlt iii`B. (b)`iiigt ivgt igt ii`C. (c )`ilt iilt ivlt iii`D. (d)`ilt ivlt iiilt ii` |
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Answer» Correct Answer - a At. Wt. of `P=31` and atomicity of `P` in while `P` is `4 Mol`. Wt. of white `P=31xx4=124` `:. 6.023xx10^(23)` molecules of white `P` weight `124 g` `:. 3.0115xx10^(23)` molecules of white `P` weight `124/6.023xx3.015` `=62 g` (ii) Wt. of 1 mole of `H_(2) gas =2 g` Wt. of 10 moles of `H_(2) gas =2xx10=20 g` (iii) `1 g` molecule of anhydrous `Na_(2)CO_(3)` = Mol wt. of `Na_(2)CO_(3)` in `g=106 g` (iv) At STP 22.4 L `CO_(2)` weight `44 g` `:. 33.6 L CO_(2)` weighs `44/22.4xx33.6 g` `66 g` So, the correct choric `ilt iilt ivlt iii` |
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| 811. |
How many grams of concentrated nitric acid solution should be used to prepare `250 mL` of `2.0 M HNO_(3)`? The concentrated acid is `70% HNO_(3)`:A. (a)`90.0` g conc. `HNO_(3)`B. (b)`70.0` g conc. `HNO_(3)`C. (c )`54.0` g conc. `HNO_(3)`D. (d)`45.0` g conc. `HNO_(3)` |
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Answer» Correct Answer - d `1000 mL` has 2 moles `250 mL` has `2/1000xx250=0.5` moles `implies n=W/M implies W=nxxM-0.5xx63 g` As `70 g` of `HNO_(3)` is in `100 g` solution So, `0.5xx63 g` of `HNO_(3)` is in `100/70xx0.5xx63=45 g` [We have assumed `70%` (mass by mass), although it should have been reported.] |
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| 812. |
`KMnO_(4)` react with oxalic acid according to the equation, `2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+) rarr 2Mn^(2+)+ 10 CO_(2)+8H_(2)O`, here `20 ml` of `0.1 M KMnO_(4)` is equivalemt toA. 20 ml of 0.5 M `H_(2)C_(2)O_(4)`B. 50 ml of 0.1 M `H_(2)C_(2)O_(4)`C. 50 ml of 0.5 M `H_(2)C_(2)O_(4)`D. 20 ml of 0.1 M `H_(2)C_(2)O_(4)` |
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Answer» Correct Answer - B `KMnO_(4)` Oxialic acid `(M_(1)V_(1))/n_(1)=(M_(2)V_(2))/n_(2), (20xx0.1)/2=(M_(2)V_(2))/5, M_(2)V_(2)=5` for (b) `M_(1)V_(1)=50xx0.1=5` |
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| 813. |
The volume of 0.1 M `Ca (OH)_(2)` needed for the neutralization of 40 ml of 0.05 M oxalic acid isA. 10 mlB. 20 mlC. 30 mlD. 40 ml |
| Answer» Correct Answer - B | |
| 814. |
`224 mL` of a triatomic gas weights `1 g at 273K` and `1 atm`. The mass of one atom of this gas is:A. `8*30xx10^(-23) g`B. `2*08xx10^(-23)` gC. `5.53xx10^(-23)` gD. `6.24xx10^(-23)` g |
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Answer» Correct Answer - C 224 mL of gas `X_(3)` has mass = 1 g `:.` 22,400 mL of gas has mass `=(1)/(224)xx22,400 = 100g` Mass of one atom `= (100)/(3xx6.02xx10^(23))` `= (16.61xx10^(-23))/(3)g = 5.53xx10^(-23)g` |
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| 815. |
Calculate the number of oxalic acid molecules in `100 mL` of `0.02 N` oxalic acidA. (a)`6.023xx10^(20)`B. (b)`6.023xx10^(21)`C. (c )`6.023xx10^(22)`D. (d)`6.023xx10^(23)` |
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Answer» Correct Answer - a Normality = Molarity`xx` Valence factor `:.` Molarity`="Narmality"/("Valence factor")` Valence factor for oxalic acid `=("mo. Wt. of oxalic acid")/("Eq. wt. of oxalic acid")` `=("mo. Wt. of oxalic acid")/("mo. Wt. of oxalic acid")xx 2` (As basically of oxalic acid is) `"Molarity"=20/2=0.01` Number of millimoles `=0.01xx100` Number of moles=`0.001` `:.` Number of oxalic acid molecules`=0.001xx6.023xx10^(23) = 6.023xx10^(20)` |
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| 816. |
If 1 ml of water contains 20 drops. Then no. of molecules in a drop of water isA. `6.023xx10^(23)` moleculesB. `1.376xx10^(26)` moleculesC. `1.344xx10^(18)` moleculesD. `4.346xx10^(20)` molecules |
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Answer» Correct Answer - C 22400 ml of water contains `=6.023xx10^(23)` molecules 1 ml of water contains `=(6.023xx10^(23))/22400` molecules `=20` drops `:.` 1 drop of water will contain `=(6.023xx10^(23))/(224xx2xx10^(3))` `=1.344xx10^(18)` molecules. |
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| 817. |
The number of atoms in 0.1 mol of a triatomic gas is:A. (a)`6.026xx10^(23)`B. (b)`1.806xx10^(23)`C. (c )`3.600xx10^(23)`D. (d)`1.80xx10^(23)` |
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Answer» Correct Answer - b Let gas be `X_(3)` Then `X_(3) -= 3X` 1 mol of `X_(3)` has 3 moles of `X` atoms 0.1 mol of `X_(3)` has `0.3` moles of `X` atoms That is `0.3xxN_(A)` 0.1 mole has atoms `=0.1xx6.02xx10^(23)xx3` `=1.806xx10^(23)` |
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| 818. |
The number of molecules in 100 mL of 0.02 `NH_(2)SO_(4)` is:A. `6.02xx10^(22)`B. `6.02xx10^(21)`C. `6.02xx10^(20)`D. `6.02xx10^(18)` |
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Answer» Correct Answer - C 0.02 N `H_(2)SO_(4)=0.01M H_(2)SO_(4)` 1000 mL of 0.001 M `H_(2)SO_(4)` contains `H_(2)SO_(4) = 0.01` mole 100 mL of 0.01 M `H_(2)SO_(4)` contains `H_(2)SO_(4) = 0.001=10^(-3)` mole Molecules present `=10^(-3)xx6.02xx10^(23)=6.02xx10^(20)` |
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| 819. |
If 1 ml of water contains 20 drops. Then no. of molecules in a drop of water isA. `6.023xx10^(23)`B. `1.376xx10^(26)`C. `1.673xx10^(21)`D. `4.346xx10^(20)` |
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Answer» Correct Answer - C 1 mL of water = 1 g of water `:.` 20 drops of `H_(2)O = 1g` 1 drop of `H_(2)O = (1)/(20)g=0.05g` `=(0.05)/(18)`mole `=(0.05)/(18)xx6.23xx10^(23)` molecules `=1.673xx10^(21)` molecules. |
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| 820. |
`25.3 g` sodium carbonate, `Na_(2)CO_(3)`, was dissolved in enough water to make `250 mL` of solution. If sodium carbonate dissociates completely, molar concentration of `Na^(+)` and carbonate ions are respectively:A. (a)`0.9555 M and 1.910 M`B. (b)`1.910 M and 0.955 M`C. (c )`1.90 M and 1.090 M`D. (d)`0.477 M and 0.477 M` |
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Answer» Correct Answer - b `M_(Na_(2)CO_(3))=M_(Na^(+))xx2=M_(CO_(3)^(2-)` and `M_(Na_(2)CO_(3))` `=(25.3xx1000)/(106xx250)=0.955` `implies 0.955xx2=1.91 M` Thus (b) is correct. |
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| 821. |
`108÷7.2=14.583`. The correct answer to this problem in proper number of significant digits isA. `15`B. `14.58`C. `14.5`D. None of these |
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Answer» Correct Answer - A `{:(108,÷7.2,=14.5833),("Three S.F. ","Two S.F.",):}` Here answer should have two significant figures. Therefore, correct answer, after rounding off two significant figures is 15. |
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| 822. |
In an experiment, 4g of `M_(2)O_(x)` oxide was reduced to 2.8g of the metal. If the atomic mass of the metal is `56g"mol"^(-1)`, the number of oxygen atoms in the oxide is:A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C `M_(2)O_(x) overset("Reduction")(rarr) M` Eq. of `M_(2)O_(x)=` eq. of Metal `("wt. of "M_(2)O_(x))/("Eq. wt. of "M_(2)O_(x))=("Wt. of Metal")/("Eq. wt. of Metal")` `4/((2xx56+xxx16)/(2x))=2.8/(56/x)` ...(i) On solving we get, `4/(56+8x)=2.8/56 implies 1/(14+2x)=1/20 implies 2x=6 implies x=3` Hence, the oxide is `M_(2)O_(3)`. |
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| 823. |
A mixture of `1.65 xx 10^(21)` molecules of X and `1.85 xx 10^(21)` molecules of Y weighs 0.688 g . The mol . wt . Of X is (Assume mol . wt . Of Y is 187)A. 52B. 84C. 126D. 41.47 |
| Answer» Correct Answer - D | |
| 824. |
Vapour density of mixture of `NO_(2)` and `N_(2) O_(4)` is 34.5 , then percentage abundance of `NO_(2)` in mixture isA. `50%`B. `25%`C. `40%`D. `60%` |
| Answer» Correct Answer - A | |
| 825. |
A mixture of `FeO` and `Fe_(3)O_(4)` when heated in air to a constant weight, gains 5% of its weight. Find the composition of the intial mixutre.A. 79 . 75B. 20 . 25C. 30 . 25D. 10 . 25 |
| Answer» Correct Answer - B | |
| 826. |
How many mole of magnesium phosphate `Mg_(3)(PO_(4))_(2)` will contain `0.25 "mole"` of oxygen atoms?A. (a)`2.5xx10^(-2)`B. (b)`0.02`C. (c )`3.125xx10^(-2)`D. (d)`1.25xx10^(-2)` |
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Answer» Correct Answer - C `8 "mole" O-=1 "mole" Mg_(3)(PO_(4))_(2)` `implies 0.25 "mole"implies1/8xx0.25` `=3.125xx10^(-2)` moles |
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| 827. |
Which of the following pairs have the same number of atoms?A. 16 of `O_(2)(g)` and 4 g of `H_(2)(g)`B. 16 g of `O_(2)` and 44 g of `CO_(2)`C. 28 g of `N_(2)` and `32g` of `O_(2)`D. 12 g of `C(s)` and 23 g of `Na(s)` |
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Answer» (c) Number of atoms isn 28g of `N_(2)=(28)/(28)xxN_(A)xx2=2N_(A)` (whre, `N_(A)`=Avogadro number) Number of atoms in 32 g of `O_(2)=(32)/(32)xxN_(A)xx2=2N_(A)` (d) 12 g of C(s) contains atoms `=(12)/(12)xxN_(A)xx1=N_(A)` Number of atoms in 23g of Na(s)`=(23)/(23)xxN_(A)xx1=N_(A)` |
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| 828. |
Calculate the total number of atoms in 2 litre gaseous mixture of `CH_(4)` and `SO_(2)` at STP which have vapour density 26. |
| Answer» Correct Answer - `(6.5)/(22.4)N_(A)` | |
| 829. |
Specific volume of cylindrical virus particle is `6.02xx10^(-2) c c//g` whose radius and length `7 Å` and `10 Å` respectively. If `N_(A)=6.02xx10^(23)`, find molecular weight of virus:A. (a)`3.08xx10^(3) kg//mol`B. (b)`15.4 kg//mol`C. (c )`1.54xx10^(4) kg//mol`D. (d)`3.08xx10^(4) kg//mol` |
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Answer» Correct Answer - B Specific volume `("volume of 1 g")` of cylindrical virus particle `=6.02xx10^(-2)` "c c/g" Radius of virus (r )`=7Å=7xx10^(-8) cm` Length of virus`=10xx10^(-8)cm` Volume of virus `pir^(2)=22/7xx(7xx10^(-8))^(2)xx10xx10^(-8)=154xx10^(-23)` c c Wt. of one virus particle`=("volume")/("specific volume")` `:.` Mol. wt. of virus=wt. of `N_(A)` particle `=(154xx10^(-23))/(6.02xx10^(-2))xx6.02xx10^(23)` `=15400 g//mol=15.4 kg//mol` |
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| 830. |
Which of the following pairs contain equal number of atoms?A. 22.4 L (STP) of nitrous oxide and 22.4 L of nitrifc oxideB. 1 millimole of HCl and 0.5 millimole of `H_(2)S`C. 1 mole of `H_(2)O_(2)` and 1 mole of `N_(2)O_(4)`D. 11.2 cc (STP) of nitrogen and 0.015g of nitric oxide |
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Answer» d) Number of atoms in `N_(2)` `=(11.2 xx 10^(-3) xx 6.023 xx 10^(23) x 2)/(22.4) = (6.023 xx 10^(20)` Number of atoms in NO = `(0.015 xx 2 xx 6.023 xx 10^(23))/(30)` `=6.023 xx 10^(20)` |
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| 831. |
Vapour density of a gas is 11.2. Volume occupied by 2.4 g of this at STP will beA. 2.4 LB. 2.24 LC. 22.4 LD. 11.2L |
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Answer» a) Molecular mass = 2 `xx` vapour density `=2xx 11.2 = 22.4` Number of moles of gas = (2.4)/(22.4)` `therefore` 1 mole occupies = 22.4 L volume `therefore (2.4)/(22.4)` mole will occupy `=22.4 xx (2.4)/(22.4)` |
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| 832. |
The strongest reducing agent isA. `F^(-)`B. `Cl^(-)`C. `Br^(-)`D. `I^(-)` |
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Answer» Correct Answer - D `I^(-)` act as a more reducing agent than other ions. |
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| 833. |
Strongest reducing agent isA. KB. MgC. AlD. Br |
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Answer» Correct Answer - A Potassium has higher negative value of reduction potential hence it shows more reducing properties. |
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| 834. |
Assertion (A): `HNO_(3)` acts only as an oxidising agent, while `HNO_(2)` acts both as an oxidising agnet and a reducing agent. Reason (R ): The oxidation number of `N` in `HNO_(3)` is maximum.A. Statement 1 is true, statement 2 is true, statement 2 is a correct explanation for statement 4B. Statement 1 is true, statement 2 is true, statement 2 is not a correct explanation for statement 4C. Statement 1 is true, statement 2 is falseD. Statement 1 is false, statement 2 is true |
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Answer» Correct Answer - A Both (A) and (2) are correct, and (2) is the correct explanation for (1) since the oxidation number of N in `HNO_(3)` is maximum (+5), therefore, it can only decrease. Hence, `HNO_(3)` acts as an oxidising agent. In `HNO_(2)`, the oxidation number of N is +3, so it can increase by losing electrons or can decrease by accepting electrons. Therefore, `HNO_(2)` acts both as an oxidising as well as a reducing agent. |
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| 835. |
Pick out the stronger reducing agent among the following oxyacids of phosphorusA. Hypophosphorous acidB. Phosphorous acidC. Hypophosphoric acidD. Pyrophosphorous acid |
| Answer» Correct Answer - A | |
| 836. |
Which of the following is the strongest oxidising agent?A. `BrO_(3)^(-)//Br^(2+), E^(@)= +1.50`B. `Fe^(3+)//Fe^(2+), E^(@)= + 0.76`C. `MnO_(4)^(-)//Mn^(2+), E^(@)=+1.52`D. `Cr_(2)O_(7)^(2-)//Cr^(3+), E^(@)=+1.33` |
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Answer» Correct Answer - C Higher is the reduction potential stronger is the oxidising agent. Hence in the given options, `MnO_(4)^(-)` is strongest oxidising agent. |
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| 837. |
What is the oxidising agent in chlorine waterA. `HCl`B. `HClO_(2)`C. `HOCl`D. None of these |
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Answer» Correct Answer - C `Cl_(2)+H_(2)O rarr HCl+HOCl` `HOCl tatt HCl +[O]` `HOCl` can furnish, nascent oxygen. |
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| 838. |
Of the four oxyacids of chlorine the strongest oxidising agent in dilute aqueous solution isA. `HClO_(4)`B. `HClO_(3)`C. `HClO_(2)`D. `HOCl` |
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Answer» Correct Answer - A `HClO_(4)` is the strongest oxidising agent. The correct order of oxidising power is `overset(+1)(HClO) lt overset(+3)(HClO_(2)) lt overset(+5)(HClO_(3)) lt overset(+7)(HClO_(4))`. |
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| 839. |
At `300 K` and `1 atm, 15 mL` of a gaseous hydrocarbon requires `375 mL` air containing `20% O_(2)` by volume for complete combustion. After combustion, the gases occupy `330 mL`. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon isA. `C_(3)H_(8)`B. `C_(4)H_(8)`C. `C_(4)H_(10)`D. `C_(3)H_(6)` |
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Answer» `C_(x)H_(y)(g)+underset(75 mL)((x+(y)/(4)))O_(2)(g)rarrxCO_(2)underset(30 mL)((g))+(y)/(2)H_(2)O(l)` `O_(2)` used `= 20%` of `375 = 75 mL` Inert part of air `= 80%` of `375 = 300 mL` Total volume of gases `= CO_(2)`+ Inert part of air `= 30 + 300 = 330 mL` `(x)/(1) = (30)/(15) rArr x = 2` `(x+y)/(1) = (75)/(15) rArr x+(y)/(4) = 5` `rArr x = 2, y = 12 rArr C_(2)H_(12)` |
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| 840. |
The molecular formula of a commercial resin used for exchanging ions in water softening is `C_(8)H_(7)SO_(3)Na(mol. Wt. 206)` . What would be the maximum uptake of `Ca^(2+)` ions by the resin when expressed in mole per gram resin?A. `(1)/(103)`B. `(1)/(206)`C. `(2)/(309)`D. `(1)/(412)` |
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Answer» Correct Answer - D We know the molecular weight of `C_(8)H_(7)SO_(3)Na` `=12 xx 8 + 1xx7 + 32+16 xx 3 + 23 = 206` we have to find, mole per gram of resin. `:. 1g` of `C_(8)H_(7)SP_(3)Na` has number of mole `= ("weight of given resin")/("Molecular, weight of resin") = (1)/(206)mol` Now, reaction looks like `2C_(8)H_(7)SO_(3)Na+Ca^(2+) rarr (C_(8)H_(7)SO_(3))_(2)Ca+2Na` `because` 2 moles of `C_(8)H_(7)SO_(3)Na` will combine with `(1)/(2) mol Ca^(2+)` `:. (1)/(2)` mole of `C_(8)H_(7)SO_(3)Na` will combine with `(1)/(2) xx (1)/(206) mol Ca^(2+) = (1)/(412)mol Ca^(2+)` |
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| 841. |
50 mL of a solution containing 1 g each of `Na_(2)CO_(3), NaHCO_(3)` and NaOH was titrated with N-HCl. What will be the titre reading when only phenolphalein is used as indicator?A. (a)35 mLB. (b)32.5 mLC. (c )24.5 mLD. (d)34.4 mL |
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Answer» Correct Answer - d At the end point using phenolphthalein as indicator uses complete `NaOH` and `1/2` Meq. Of `Na_(2)CO_(3)`. `:.` Meq. of `NaOH+1/2` meq. of `Na_(2)CO_(3)`= Meq. Of HCl `1/40xx1000+1/2xx1/106xx2xx1000=Vxx1` `:. V=34.4 ml` |
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| 842. |
`40 mL 0.05 M` solution of sodium sesquicarbonate dehydrate `(Na_(2)CO_(3).NaHCO_(3).2H_(2)O)` is titrated against `0.05 M HCl` solution, `x mL` of acid is required to reach the phenolphthalein end point while mL of same acid were required when methyl organe indicator was used in a separate titration. Which of the following is (are) correct statements?A. `y-x = 80 mL`B. `y+x = 160 mL`C. If the titration is started with phenolphthalein indicator and methyl orange is added at the end point, `2 x mL` of `HCl` would be required further to reach the end pointD. If the same volume of same solution is titrated against `0.10 M NaOH, x//2 mL` of base would be required |
| Answer» Correct Answer - A::B::C::D | |
| 843. |
When aqueous solution of `Na_(2)S` is titrated with dilute and acidified `KMnO_(4)` solution, `Na_(2)SO_(3)` is formed. In this reaction, moles of `KMnO_(4)` reduced per mole of `Na_(2)S` isA. `0.833`B. `1.2`C. `1.50`D. `1.8` |
| Answer» Correct Answer - B | |
| 844. |
Which one of the following does not get oxidised by bromine waterA. `Fe^(+2)` to `Fe^(+3)`B. `Cu^(+)` to `Cu^(+2)`C. `Mn^(+2)` to `MnO_(4)^(-)`D. `Sn^(+2)` to `Sn^(+4)` |
| Answer» Correct Answer - C | |
| 845. |
in neutral or faintly alkaline solution, 8 moles of permanganate anion quantitatively oxidize thiosulphate anions to produce X moles of a sulphur containing product. The magnitude of X is |
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Answer» Correct Answer - 6 `8MnO_(4)^(-)+3 S_(2)O_(3)^(2-)+H_(2)O overset("Faintly alkaline")(rarr) 8 MnO_(2)+6 SO_(4)^(2-) +2OH^(-)` `:.` 8 moles `MnO_(4)^(-)` produce 6 moles `SO_(4)^(2-)` |
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| 846. |
In dilute aqueous `H_(2)SO_(4)` the complete diaquadioxalatoferrate (II) is oxidised by `MnO_(4)^(-)`. For the reaction, the ratio of the rate of change of `[H^(+)]` to the rate of change of `[MnO_(4)^(-)]` is |
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Answer» Correct Answer - 8 `8H^(+)+MnO_(4)^(-)+[Fe(H_(2)O)_(2)(OX)_(2)]^(2-) rarr Mn^(2+)+Fe^(3+)+4CO_(2)+6 H_(2)O` `("rate of change of "[H^(+)])/("rate of change of "[MnO_(4)^(-)])=8` |
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| 847. |
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 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 200mL of hydrogen will combine with 100mL oxygen to produce 200mL of water vapour |
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Answer» The element carbon, combines with oxygen to form two compounds namely carbon dioxide and carbon m onoxide. In `CO_(2),12` parts by mass of carbon combine with 32 parts by mass of oxygen while in CO,12 parts by mass of carbon combine with 16 parts by mass of oxygen. Therefore, the masses of oxygen combine with a fixed mass of carbon (12parts) in `CO_(@)` and CO are 32 and 16 respectively. These masses of oxygen bear a simple ratio of 32:16of 2:1 to each other This is an example of law of multiple proportion. |
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| 848. |
Which of the following statements indicates that law of multiple proportion is being followed?A. Sample of carbon dioxide taken form any source will always have carbon and oxygen in the ratio 1 : 2B. Carbon forms two oxides namely `CO_(2)`and CO, where masses of osygen 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 wqual to theD. At constant temperatur in magnesium oxide formed hydrogen will combine with 100 mL oxygen to produce 200 mL of water vapour |
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Answer» Correct Answer - B The element, carbon combines with oxygen to from two compounds, namely, carbon dioxide and carbon monoxide. in `CO_(2),` 12 parts by mass of carbon combine with 32 parts by mass of oxygen while in CO, 12 parts by mass of carbon combine with 16 parts by mass of oxygen. Therefore, the masses of oxygen combine with a fixed mass of carbon (12 parts) in `CO_(2)` and CO are 32 and 16 respectibely. These masses of oxygen bear a simple ratio of 32 : 16 or 2 : 1 to each other This is an example of law of multiple proportion. |
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| 849. |
The law of multiple proportion is illustrated byA. carbon monoxide and carbon dioxideB. potassium bromide and potassium chlorideC. water and heavy waterD. calcium hydroxide and barium hydroxide |
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Answer» Correct Answer - A Law of multiple proportions deals with two elements forming more than one compound. |
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| 850. |
A solution is made by dissolving 49g of `H_2SO_4` in 250 mL of water. The molarity of the solution prepared isA. 2 MB. 1 MC. 4 MD. 5 M |
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Answer» Molarity `=("Wt. of solute")/("Mol. Wt. of solute")xx(1000)/("Volume of soln. (mL)")` =`49/98xx1000/250` =2M |
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