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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.
| 601. |
Calculate the number of atoms of each type that are present in `3.42 g` of sucrose `(C_(12) H_(22) O_(11))`. |
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Answer» `Mw` of sucrose `= C_(12) H_(22) O_(11)` `= 12 xx 12 + 1 xx 22 + 16 11` `= 342 g` a. 1 mol of sucrose `= 342 g = 12 mol of C` `:. 3.42 g = (12)/(342) xx 3.42` `= 0.12 mol of C = 0.12 xx 6.023 xx 10^(23) "atoms"` `= 7.228 xx 10^(22) "atoms of" C` b. `342 g` of Sucrose `= 22 mol of H` `3.42 g` of sucrose `= (22)/(342) xx 3.42` `= 0.22 mol of H` `= 0.22 xx 6.023 xx 10^(23) `atoms `=132.5 xx 10^(21)` atom of `H` c. `342 g` of sucrose `= (11 xx 6.023 xx 10^(23))/(342) xx 3.42` `= 66.25 xx 10^(21)` atoms of `O` |
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| 602. |
Weight of copper oxide obtained by heating `2.16 g` of metallic copper with `HNO_(3)` and subsequent ingnition was `2.70 g`. In another experiment, `1.15 g` of copper oxide on reduction yielded `0.92 g` of copper. Show that the results illustrate the law of definite proportions. |
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Answer» In the first experiment Weight of copper `= 2.16 g` Weight of copper oxide `= 2.70 g` Weight of oxygen `= 2.70 - 2.16 = 0.54 g` Percentage of copper `= ("Weight of copper")/("Weight of copper oxide")xx100=((2.16g))/((2.70g))xx100=80%` Percentage of oxygen `= ("Weight of oxygen")/("Weight of copper oxide")xx100=((0.54g))/((2.70g))xx100=20%` In the second experiment Weight of copper oxide `= 1.15 g` Weight of copper `= 0.92 g` , Weight of oxygen `= 0.23 g` Percentage of copper `= ("Weight of copper")/("Weight of copper oxide")xx100=((0.92g))/((1.15g))xx100=80%` Percentage of oxygen `= ("Weight of oxygen")/("Weight of copper oxide")xx100=((0.23g))/((1.15g))xx100=20%` Since the ratio by weights of copper and oxygen in the two compounds remains the same, the law of definite composition is illustrated. |
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| 603. |
Weight of copper oxide obtained by heating `2.16 g` of metallic copper with `HNO_(3)` and subsequent ingnition was `2.70 g` In another experient, `1.15 g` of copper oxide on reduction yielded `0.92 g` of copper. Show that the results illustrance the law of definite proportions. |
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Answer» Case I : `underset((2.16 g))(Cu) + HNO_(3) rarr Cu(NO_(3))_(2) overset(Delta)rarr underset(2.7 g)(CuO)` Weight of oxygen `= 2.7 - 2.16 = 0.54 g` % of `O_(2)` in `CuO = (0.54 xx 100)/(2.7) = 20%` case II : Copper oxide taken `= 1.15 g` Copper left `= 0.92 g` oxygen present `= 1.15 - 0.92 = 0.23 g` % of `O_(2)` in `CuO = (0.23 xx 100)/(1.15) = 20%` |
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| 604. |
In `a` solution the concentration of `CaCl_(2)` is `5M &` that of `MgCl_(2)` is `5 m`. The specific gravity of solution is `1.05`, calculate the concentration of `Cl^(-)` in the solution in terms of Molarity. |
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Answer» Correct Answer - `[Cl^(-)]=13.36M` `CaCl_(2)rarr5M=555g` in `1L` solution or in `1050g` solution `wt`. Of `("solvant" + MgCl_(2))=1050-555=495g` `MgCl_(2)rarr5m` `1000g` solvant `rarr 5` mol of `MgCl_(2)` `=5xx95=475gMgCl_(2)` `i.e., 1475 ("solvant" + MgCl_(2))rarr 475g MgCl_(2)` `495("solvant"+ MgCl_(2)) rarr(475)/(1475)xx495` `=159.4g MgCl_(2)` moles of `MgCl_(2)=(159.4)/(95)=1.678` Total moles of `Cl^(-)` `=(5+1.678)xx2=13.356` volume of solution `1L` Molarity of `Cl^(-) = 13.356 M` |
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| 605. |
`100 g` sample of clay (containing 19% `H_(2)O`, 40% silica, and inert inpurities as rest) is partically dried so as to contains 10% `H_(2) O`. Which of the following `is//are` correct statements (s) ?A. The percentage of silica in it is 44.4%B. The mas of partically dried clay is `90.0 g`.C. The precentage of inert impurity in it is 45.6%D. The mass of water evaporated is `10.0 g` |
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Answer» Correct Answer - A::C `underset({:(%"in original clay"implies),(%"after partial drying"implies):})( ) underset({:(40),(a):})("Silica") underset({:(19),(10):})(H_(2)O) underset({:(100 - (40 + 19) = 41),(100 - (a + 10) = 90 - a):})("Impurities)` On heating, only water evaporates from clay, whereas silica and impurities are left as it is. Therefore, % ratio of silica and impurities remains unchanged, i.e., `(40)/(a) = (41)/(90 - a), :. a = 44.4%` % of mipurities after partial drying `= (19 - 10) = 9 g` `= 45.6%` Mass of `H_(2) O` evaporated `= (19 - 10) = 9 g` |
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| 606. |
A partially dried clay mineral contains `8%`d water. The original sample contained `12%` water and `45%` sillica. The `%` if sillica in the partially dried sample is nearly:A. `50%`B. `49%`C. `55%`D. `47%` |
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Answer» Correct Answer - 4 Let original sample weighs `xx g` water `=(12)/(100)x, ` silica `=(45)/(100)xxx` Impurities `=(43)/(100)x` Partially dried sample weight `y g` Water `=(8)/(100)y` Since no evaporational silica `&` impurites. So, `(45)/(100)x+(43)/(100)x=(92)/(100)yrArrx=(92)/(88)y` `%` of silica in partially dried sample `((45)/(100)xxx)/(y)xx100=47%` |
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| 607. |
For the reaction `2P + Q rarr R, 12` mol of P and 8 mol of Q are taken then:A. 3 mol of R is producedB. 6 mol of R is producedC. 25% of Q is left behindD. 25% of Q has reacted |
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Answer» Correct Answer - B::C |
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| 608. |
Calculate moles of electrons in 1900 mg of `PO_(4)^(3-)` ion. |
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Answer» Correct Answer - 1 |
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| 609. |
A regular cube of metal measures exactly 10cm on an edge and has density `8g/c c`. If the cube contains `6 xx 10^(25)` atoms of the metal, determine atomic weight of metal? (Take `N_(A) = 6 xx 10^(23)`)A. 40B. 60C. 80D. 100 |
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Answer» Correct Answer - C |
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| 610. |
A crystalline polymer molecule is uniform prismatic in shape with dimension as shown. If density of the polymer is `1.2g/cm^(3)`. Find its molar mass. A. `939 xx 10^(3)`B. `939 xx 10^(-3)`C. `632 xx 10^(3)`D. Insufficient data |
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Answer» Correct Answer - A |
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| 611. |
The mass of 70% `H_(2)SO_(4)` required for neutralization of one mole of NaOH is:A. 70gB. 35gC. 30gD. 95gm |
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Answer» Correct Answer - A |
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| 612. |
A drug marijuna owes its activity to tetrahydrocarbinol, which contains 70% as many C atoms as oxygen atoms. The number of mole of compound in a gm of it is 0.00318. the molecualr formular will be :A. `C_(20)H_(30)O_(2)`B. `C_(21)H_(30)O_(2)`C. `C_(12)H_(20)O_(2)`D. `C_(12)H_(20)O_(3)` |
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Answer» Correct Answer - B |
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| 613. |
Valency factor of the following compounds will be same in neutralisation -A. `SO_(2) rarr H_(2)SO_(3)`B. `NH_(3)+H^(+)rarr NH_(4)^(+)`C. `N_(2)O_(6) rarr 2HNO_(3)`D. `A` and `C` have same valency factor |
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Answer» Correct Answer - 4 `(1) SO_(2)rarr H_(2)SO_(3)" "((M)/(E)=2)` `(2)NH_(3)+H^(+)rarr NH_(4)^(+)" "((M)/(E)=1)` `(3) N_(2)O_(5)rarr 2HNO_(3)" "((M)/(E)=2)` |
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| 614. |
Assertion : During a chemical reaction, total moles remains constant. Reason: During a chemical reaction, total mass remains constant.A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1C. Statement-1 is false, statement -2 is true.D. Statement -1 is true, statement-2 is false. |
| Answer» Correct Answer - D | |
| 615. |
40 gm of a carbonate of an alkali metal or alkaline earth metal containg some insert impurities was made to react with excess HCl solution. The liberated `CO_(2)` occupied 12.315 litre at 1 atm and 300K. The corrrect option is:A. Mass of impurity of 1gm and metal is BeB. Mass of impurity is 3gm and metal is LiC. Mass of impurity is 6gm and metal is LiD. Mass of impurity is 2gm and metal is Mg |
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Answer» Correct Answer - B |
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| 616. |
40 gm of a carbonate of an alkali metal or alkaline earth metal containing some insert impurities was made to react with excess HCl solution. The liberated `CO_(2)` occupied 12.315 lit. at 1 atm & 300 K . The correct option isA. Mass of impurity is 1 gm and metal is BeB. Mass of impurity is 3 gm and metal is LiC. Mass of impurity is 5 gm and metal is BeD. Mass of impurity is 2 gm and metal is Mg |
| Answer» Correct Answer - B | |
| 617. |
1 mole of `H_(2)SO_(4)` will exactly neutralise:A. 2 mole of ammoniaB. 1 mole of `Ba(OH)_(2)`C. 0.5 mole of `Ca(OH)_(2)`D. 2 mole of KOH |
| Answer» Correct Answer - A::B::D | |
| 618. |
Hexamethylenediamine `[H_(2)N(CH_(2))NH_(2)]` reacts with adipic acid `[HOoverset(O)overset(||)C-(CH_(2))_(4)-overset(O)overset(||)(C)-OH]` to form dimer as: The dimer polymerises to form Nylon 6,6 as per the reaction Select the correct statement:A. 290 gm of Hexamethylenediamine is required to make 610 gm of dimer.B. 730 gm of adipic acid is required to make 610 gm of dimer.C. In order to obtain 1.03 kg of Nylon-6,6 at least 1220 gm of dimer is required.D. In order to obtain 1.13 kg of Nylon-6,6 at least 1742.9 gm of dimer is required. |
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Answer» Correct Answer - B::D |
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| 619. |
A Solution of `C_(2)H_(5)OH` and water contains 54% water by mass, then which option(s) are correct for the given solution: [Given density of solution =1 gm/ml]A. Molality =18.52B. %(w/w)=46C. `X_(C_(2)H_(5)OH)=0.25`D. %(w/w) =46 |
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Answer» Correct Answer - A::B::C::D |
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| 620. |
A fresh `H_(2)O_(2)` solution is labelled `11.2V`. This solution has the same concentration as a solution which isA. `3.4%( wt // wt)`B. `3.4% ( vol //vol)`C. `3.4%(wt // vol)`D. None of these |
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Answer» Correct Answer - 3 `V=11.2M` `rArr M=1 mol//L` Hence there are `34g H_(2)O_(2)` in 1 litre `i.e. 3.4g` in `100 mL" "rArr" "` Hence `3.4%(wt.)/(vol.)` |
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| 621. |
12 g of Mg was burnt in a closed vessel containing 32 g oxygen , which of the following is/are correct.A. 2 gm of Mg will be lift unburnt.B. 0.75 gm-molecules of `O_(2)` will be left unreacted .C. 20 gm of MgO will be formed.D. The mixture at the end will weight 44 g. |
| Answer» Correct Answer - B::C::D | |
| 622. |
50 gm of `CaCO_(3)` is allowed to react with 68.6 gm of `H_(3)PO_(4)` then select the correct option(s): `3CaCO_(3) +2H_(3)PO_(4) rarr Ca_(3)(PO_(4))_(2)+3H_(2)O+3CO_(2)`A. 51.67 gm salt is formedB. Amount of unreacted reagent =35.93 gmC. `n_(Co_(2))` =0.5 molesD. 0.7 moles of `CO_(2)` is evoled |
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Answer» Correct Answer - A::B::C |
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| 623. |
`50 g` of `CaCO_(3)` is allowed to react with `70 g` of `H_(3)PO_(4^(.)` Calculate : `(i)` amount of `Ca_(3)(PO_(4))_(2)` formed `(ii)` amount of unreacted reagent |
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Answer» Correct Answer - `(i)51.66g" "(ii)37.31g` `3CaCO_(3)+2H_(3)PO_(4)rarrCa_(3)(PO_(4))_(2)+3H_(2)O+3CO_(2)` `{:(50//100"mole",70//98"mole"),(=0.5,0.7142),(" "-,0.7142-(2)/(3)xx0.5=0.3808((0.5)/(3))):}` Limiting reactant `m_(CaCO_(3))=(0.5)/(3)xxM_(Ca_(3)(PO_(4))_(2))=51.66gm` `m_(H_(3)PO_(4))=0.3808xxM_(H_(3)PO_(4)=31.31gm` |
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| 624. |
50 gm of `CaCO_(3)` is allowed to react with 68.6 gm of `H_(2)PO_(4)` then select the correct option(s)- `3CaCO_(3)+ 2H_(3)PO_(4) to Ca_(3)(PO_(4))_(2) + 3H_(2)O + 3CO_(2)`A. 51.67 gm salt is formedB. Amount of unreacted reagent =35.93 gmC. `n_(CO_(2))`=0.5 molesD. 0.7 moles `CO_(2)` is evolved |
| Answer» Correct Answer - A::B::C | |
| 625. |
In an experiment to verify the value of absolute zero, a student is instructed to measure the volume of He in a 10mL syrings at `10^(@)C`. She is told to plot the volume versus temperature and to extrapolate this graph to zero volume and read the resulting temperature. Which modification of the experimental procedure will give the best value for absolute zero?A. Correcting each measured volume to one atmosphere pressure before plotting.B. Doubling the number of temperature-volume value between `0^(@)C` and `100^(@)C`.C. using a thermometer that can measure temperature to `+-O^(@)C` between `O^(@)C` and `100^(@)C`.D. Measuring the volume of He in the syringe at `-40^(@)C` and `80^(@)C`. |
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Answer» Correct Answer - D |
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| 626. |
`4.08 g` of a mixture of `BaO` and an unknown carbonate `MCO_(3)` was heated strongly. The residue weighed `3.64 g`. This was dissolved in `100 mL` of `1 N HCl`. The excess of acid required of `16 mL` of `2.5 N NaOH` for complete neutralisation. Identify the matal `M`. |
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Answer» On heating, `BaO` does not decompose. Only corbonate decomposes as the follows. `MCO_(3) rarr MO (s) + CO_(g)` `1 mol of MCO_(3) -= 1 mol of MO` `-= 1 mol of CO_(2)` Los in weight on heating is due to the loss of `CO_(2)`. Weight of `CO_(2) = 4..08 - 3.64 = 0.44 g -= 0.01 "mole" of CO_(2)` From stochiometry, Moles of `MO - 2 (0.01)` (`n` factor `= 2`) Let `A =` atomic mass of `M` Molecular mass of `MCO_(3) = a + 60` `implies` mass of `BaO = 4.08 - 0.01 (A + 60)` Now both the oxides, `BaO` and `MO`, will react with `HCl`. At neutralisation stage: mEq of oxides = mEq of `HCl` mEq of `HCl` used for oxides `= 1 xx 100 - 2.6 xx 16 = 60` `implies mEq of MO + mEq of BaO = 60` `implies [2(0.01) + (4.08 - 0.01 (A + 60))/(154//2)] xx 1000 = 60` On simplifying, we get: `A = 40` Hence, the metal `M` is calcium `(Ca)`. |
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| 627. |
5 L of gas at STP weight 6.25 g . What is the gram molecular weight ?A. 1.25B. 14C. 28D. 56 |
| Answer» Correct Answer - 3 | |
| 628. |
`27 g` of `Al` will react completely with…… `g` of `O_(2)`A. `8 g`B. `10 g`C. `24 g`D. `49 g` |
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Answer» Correct Answer - C `4Al + 3O_(2) rarr 2Al_(2)O_(3)` `4 xx 28 g "of" Al implies 3 xx 32 g "of" O_(2)` `27 g "of" Al implies (3 xx 32)/(4 xx 27) xx 27 implies 24 g` |
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| 629. |
Calculate the mass of sucrose `C_(12)H_(22)O_(11) (s)` produced by mixing `78 g` of `C(s), 11 g` of `H_(2)(g) & 67.2 "litre"` of `O_(2)(g)` at `STP` according to given reaction (unbalanced) ?A. `171 g`B. `155.4 g`C. `185.25 g`D. None of these |
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Answer» Correct Answer - A `12 C (s)+11 H_(2) (g) + (11)/(2) O_(2) rarr C_(12)H_(22)O_(11)` `{:(,"mole",(78gm)/(12),(11)/(2),(68.1)/(27.7)),(,,=6.5,5.5,3),(,,,(LR),),(,,,,):}` mass of `C_(12)H_(22)O_(11) = 0.5 xx 342 = 171 gm` |
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| 630. |
`n-`Butance `(C_(4)H_(10))` is produced by monobromation of `C_(2)H_(6)` followed by the Wurtz reaction. Calculate the volume of ethane at `STP` requried to produce `55 g` of n-butane. The bromination takes place with 90% yield and the Wurtz reaction with 85% yield.A. `27.75 L`B. `55.5 L`C. `111 L`D. `5.55 L` |
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Answer» Correct Answer - B `2C_(2)H_(6) + Br_(2) rarr underset(underset(underset(12 xx 4 + 10 = 58g)(C_(4)H_(10)))(darr 2Na))(2C_(2)H_(5) Br + HBr)` `58 g` of `n`-butane `implies 2 xx 22.4 L "of" C_(2)H_(6)` at `STP` `implies 2 xx 22.4 xx (100)/(85) xx (100)/(90)` `58 g` of n-butane `= 58.56 L` `55 g` of n-butane `= 55.5 L` |
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| 631. |
0.6 g of carbon was burnt in the air to form `CO_(2)`. The number of molecules of `CO_(2)` introduced into the will be : `C+O_(2)toCO_(2)`A. `6.02xx10^(23`B. `3.01 xx10^(23)`C. `6.02xx10^(22)`D. `3.01xx10^(22)` |
| Answer» Correct Answer - 4 | |
| 632. |
At STP , for complete combustion of 3 g `C_(2)H_(6)` the required volume of `O_(2)` will be -A. 78. 4 LB. 7. 84 LC. 2. 78 LD. 6. 23 L |
| Answer» Correct Answer - 2 | |
| 633. |
At `25^(@)` C for complete combustion of 5 mol propane `(C_(3)H_(8))` . The required volume of `O_(2)` at STP will be .A. 560 LB. 560 mLC. 360 LD. 360 mL |
| Answer» Correct Answer - 1 | |
| 634. |
A definite amount of gaseous hydrocarbon having `("carbon atoms less than" 5)` was burnt with sufficient amount of `O_(2)`. The volume of all reactants was `600mL`. After the explosion the volume of the product `[CO_(2)(g)` and `H_(2)O (g)]` was found to be `700 mL` under the similar conditions. The molecular formula of the compound is ?A. `C_(3)H_(8)`B. `C_(3)H_(6)`C. `C_(3)H_(4)`D. `C_(4)H_(10)` |
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Answer» Correct Answer - A `{:(,C_(x)H_(y),+(x+(y)/(4))O_(2)rarr,xCO_(2)+,(y)/(2)H_(2)O),(,a,(x+(y)/(4))a,ax,(ay)/(2)),(,a, +(x+(y)/(4))a = 600,,),(,,ax+a(y)/(2)=700,,):}` `6x+3y=7+7x+7y//4` `7+x=5y//4` `x lt 5` put the value if `x=3 " " 10=5y//4 " " y=8` Ans is `C_(3)H_(8)` |
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| 635. |
Total number of electrons present in 4.4 gm oxalate ion `(C_(2)O_(4)^(-2))` is:A. `0.05N_(A)`B. `2.3 N_(A)`C. `2.2N_(A)`D. `2.1 N_(A)` |
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Answer» Correct Answer - B |
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| 636. |
A solution of stearic acid `(C_(18)H_(36)O_(2),M=284)` in benzene contains `1.42` gm acid per L. When this solution (100 L) is dropped on surface, `C_(6)H_(6)` gets eveporated and acid forms a unimolecular layer on the surface. If it covers an area 6020 `cm^(2)` with unimolecular film, find the area covered by one molecule of acid.A. `2xx10^(-20)cm^(2)`B. `4xx10^(-20)cm^(2)`C. `2xx10^(20)cm^(2)`D. `4xx10^(20)cm^(2)` |
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Answer» Correct Answer - A |
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| 637. |
The vapour density of a gas A is twice that of a gas B. If the molecular weight of B is M, the molecular weight of A will be:A. MB. 2MC. 3MD. `M/2` |
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Answer» Correct Answer - B |
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| 638. |
`6.02xx10^(20)` molecules of urea are present in 100 ml of its solution. The concentration of solution is :A. `0.001` MB. `0.01` M`C. `0.02` MD. `0.1` M |
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Answer» Correct Answer - B |
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| 639. |
How much `(NaN)_(3)` must be weighed out to make 50 ml of an aqueous solution containing 70 mg of `Na^(+)` per mL?A. `12.394` gB. `1.29` gC. `10.934` gD. `12.934` g |
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Answer» Correct Answer - D |
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| 640. |
For reaction `A+2BtoC`. The amount of C formed by starting the reaction with 5 mole of A and 8 mole of B is :A. 5 molB. 8 molC. 16 molD. 4 mol |
| Answer» Correct Answer - 4 | |
| 641. |
`2H_(2) O_(2) (l) rarr 2H_(2)o(l) + O_(2) (g)` `100 mL` of `X` molar `H_(2)O_(2)` gives `3L` of `O_(2)` gas under the condition when 1 moe occupies `24 L`. The value of `X` isA. 2.5B. 1C. 0.5D. 0.25 |
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Answer» Correct Answer - D `2H_(2)O_(2) (l) rarr 2H_(2) P (l) + O_(2) (g)` `24 L O_(2) = 1 "mol" O_(2)` `3 L O_(2) = (1)/(8) "mol" O_(2) = (1)/(4) "mol" H_(2)O_(2) "in" mL` `2.5 "mol" H_(2) O_(2) L^(-1)` |
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| 642. |
A nuggest of gold and quartz was found to contain x g of gold y g of quartz and has density d. If the densities of gold and quartz are `d_(1)` and `d_(2)` respectively, then the correct relations is:A. `x/d_(1) + y/d_(2) = (x + y)/(d)`B. `xd_(1) + yd_(2) = (x + y)d`C. `(x)/(d_(2)) + y/(d_(1)) = (x +y)/(d)`D. `(x + y)/(d) + x/d_(1) + x/d_(2) = 0` |
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Answer» Correct Answer - A |
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| 643. |
Calculate the number of atoms present in 7.1 g of chlorine ?A. `0.2 N_(A)`B. `0.3N_(A)`C. `0.4N_(A)`D. `0.5 N_(A)` |
| Answer» Correct Answer - 1 | |
| 644. |
The vapour densities of two gases are in the ratio of `1 : 3 ` . Their molecular masses are in the ratio of : -A. ` 1 : 3 `B. `1 : 2`C. `2 : 3`D. `3 : 1` |
| Answer» Correct Answer - 1 | |
| 645. |
When 40 c c of slightly moist hydrogen chloride gas is mixed with 20 c c of ammonia gas the final volume of gas left at the same temperature and pressure will be `NH_(3)(g)HCl(g)toNH_(4)Cl(s)`A. 20 c cB. 40 c cC. 60 c cD. 100 c c |
| Answer» Correct Answer - 1 | |
| 646. |
One litre of a gas at STP weight 1.16 g it can possible beA. `C_(2)H_(2)`B. COC. `O_(2)`D. `NH_(3)` |
| Answer» Correct Answer - 1 | |
| 647. |
Calculate the molarity of each of the following solutions `:` `a. 30g` of `Co(NO_(3))_(2).6H_(2)O` in `4.3L` of solution `b. 30mL` of `0.5 M H_(2)SO_(4)` diluted to `500mL`. |
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Answer» (a) Molarity of solution `= ("Mass of solute/Molar mass of solute")/("Volume of solution in litres")` Mass of solutem `Co(NO_(3))_(2).6H_(2)O=30g` Molar mass of solute, `Co(NO_(3))_(2).6H_(2)O=59+2xx14+6xx16+6xx18=291 "g mol"^(-1)` Volume of solution `=4.3 L` Molarity `(M)=((30g)//(291 "g mol"^(-1)))/((4.3L))=0.024"mol L"^(-1)=0.024M` (b) Volume of undiluted `H_(2)SO_(4)` solution `(V_(1))=30` mL Molarity of undiluted `H_(2)SO_(4)` solution `(M_(1))=0.5` M Volume of diluted `H_(2)SO_(4)` solution `(V_(2))=500` mL Molarity of diluted `H_(2)SO_(4) (M_(2))` can be calculated as : `M_(1)V_(1)=M_(2)V_(2)` or `M_(2)=(M_(1)V_(1))/(V_(2))=((30mL)xx(0.5M))/((500mL))=0.03M`. |
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| 648. |
What is the significant of N/10 NaOH solution ? |
| Answer» It means that 0.1 gram equivalent (4 g) of NaOH is dissolved per litre of the solution. | |
| 649. |
Calculate (a) Mass of `2.5` gram atoms of magnesium, (b) Gram atoms in `1.4` grams of nitrogen (Atomic mass : `Mg = 24,N=14`) |
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Answer» (a) gram atom of `Mg = 24 g` 2.5 gram atoms of `Mg = 24 xx 2.5 = 60` g (b) 1 gram atom of `N = 14 g , 14 g` of `N=1` gram atom 1.4 g of `N=(1)/(14)xx1.4=0.1` gram atom |
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| 650. |
Calculate mass of `O` atoms is `6gm CH_(3) COOH` ? |
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Answer» Molar mass of `CH_(3) COOH = 60` Moles of `CH_(3) COOH = (6)/(60) = 0.1` 1mole `CH_(3)COOH` has `2mol O` atoms `0.1` mole `CH_(3)COOH` has `0.2molO` atoms `"Mass of oxygen" = "moles" xx "Atomic mass" =0.2 xx 16 = 3.2 gm` . |
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