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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.
| 87601. |
A sample of gas changes from p_(1), V_(1) and p_(2). V_(2) and T_(2) by one path and then back to p_(1), V_(1) and T_(1). Delta U for the process is : |
| Answer» Answer :D | |
| 87602. |
A sample of gas at 35^@C and 1atmosheric pressure occupies a volume of 3.75 litre. At what temperature should the gas be kept, if it is required to reduced the volume to 3.0 litre at the same pressure |
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Answer» `0.00^@C` |
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| 87603. |
A sample of galena is contaminated with zinc blende. Name one chemical which can beused to concentrate galena selectively by froth floatation method. |
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| 87604. |
A sample of ferrous sulphate and ferrous oxalats wasdissolved in dil . H_(2)SO_(4). The complete oxidation of reaction mixture required 40 ml of N/15KMnO_(4) . After the oxidation, the reaction mixture was reduced by Zn and H_(2)SO_(4). Again on oxidation by same KMnO_(4), 25 ml " of " KMnO_(4)was required. Calculate the ratio of Fe in ferrous sulphate and oxalate. |
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| 87605. |
A sample of ferrous sulphate and ferrous oxalate was dissolved in dil. H_2SO_4. The compete oxidation of reaction mixture require 40 ml of N/15 KMnO_4. After the oxidation the reaction mixture was reduced by zn and H_2SO_4. On again oxidation by same KMnO_4 25 ml of KMnO_4 was required. Calculate the ratio of Fe atoms in ferrous sulphate and oxalate. |
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| 87606. |
A sample of ferrous oxide has actual formula Fe_(0.93)O_(1.00). In this sample what fraction of metal ions are Fe^(2+) ions ? What type of nonstoichiometric defect is present in this sample ? |
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Answer» Solution :Let number of O-atoms = 100 `therefore`Number of FE atoms = 93 Let number of `Fe^(2+)` ions = x `therefore` Number of `Fe^(3+)` ions `= (93 - x)` SINCE compound is neutral. We have `2.x + 3(93 - x) = 2 xx 100` `therefore 2x + 279 - 3x = 200` `therefore x = 79` Fraction of `Fe^(2+)` ions present `= 79/93 = 0.849` Metal deficiency defect is present in the SAMPLE because Fe are less in amount than OXYGEN required for stoichiometric proportion. |
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| 87607. |
A sample of Fe2(SO4)3 and FeC2O4 was dissolved in dil H_2SO_4. The complete oxidation of reaction mixture req. 40 ml of N/16 KMnO_4. After the oxidation mixture was reduced by Zn and dil. H_2SO_4. On again oxidation by same KMnO_4, 60 ml were require. Calculate the ratio of millimoles of Fe_2(SO_4)_3 and FeC_2O_4 in mixture. |
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| 87608. |
A sample of drinking water was found to be severely contaminated with chloroform (CHCl_(3)) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass) :(i) Express this in percent by mass.(ii) Determine the molality of chloroform in the water sample. |
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Answer» SOLUTION :(i) 15 ppm (by mass) means 15 parts per million (106) of the solution. Therefore, PERCENT by mass `= (15)/(10^(6))xx100` `= 1.5xx10^(-3)%` (ii) Molar mass of chloroform `(CHCl_(3))` (ii) Molar mass of chloroform `(CHCl_(3))` `=1xx12+1xx1+3xx35.5` `= 119.5 g mol^(-1)` Now, according to the question, 15 g of chloroform is present in 106 g of the solution. 1.e., 15 g of chloroform is present in (106 - 15) per 106 g of water : Therefore, MOLALITY of the solution `= ((15)/(119.5)mol)/(10^(6)xx10^(-3)kg)=1.26xx10^(-4)m`. |
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| 87609. |
A sample of drinking water was found to be severely contaminated with chloroform (CHCl_3),supposed to be carcinogen. The level of contamination was 15 ppm (by mass) :(i) Express this in per cent by mass.(ii) Determine the molality of chloroform in the water sample. |
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Answer» SOLUTION : 15 ppm means 15 parts in one million` (10^6) ` parts by mass in the solution ` therefore ` PERCENTAGE by mass = `(15)/(10^6) xx 100 = 15 xx 10^(-4)` Taking 15 g chloroform in `10^6` g of the solution, mass of solvent`~~10^6 g`[Neglecting 15 in COMPARISON with `10^6`] Molar mass of `CHCl_3 = 12 + 1 + 3 xx 35.5 = 119.5 g "mol"^(-1)` molality ` = (15// 119.5)/(10^6) xx 1000 = 1.25 xx 10^(-4) m` |
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| 87610. |
A sample of drinking water was found to be severely contaminated with chloroform, CHCl_(3), supposed to be carcinogen. The level of contamination was 15 ppm (by mass). (i) Express this in percent by mass (ii) Determine the molarity of chloroform in the water sample. |
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Answer» SOLUTION :(i) 15 ppm means 15 parts in million `(10^(6))` parts `therefore""%" by mass "=(15)/(10^(6))xx100=15xx10^(-4)=1.5xx10^(-3)%` (ii) Molar mass of chloroform `(CHCl_(3))=12+1+3xx35.5=118.5g mol^(-1)` `"100 g of the SAMPLE CONTAIN chloroform"=1.5xx10^(-3)g` `therefore"1000 g (1 kg) of the sample will contain chloroform "=1.5xx10^(-2)g=(1.5xx10^(-2))/(118.65)" mole"` `=1.266xx10^(-4)" mole"` `therefore"Molality "=1.266xx10^(-4)m`. |
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| 87611. |
A sample of drinking water was found to be serverly contaminated with chloroform, CHCl_(3), supposed to be carcinogen. The level of contamination was 15 ppm (by mass). (i) Express this in percent by mass. (ii) Determine the molality of chloroform in the water sample. |
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Answer» SOLUTION :15 ppm means 15 PARTS in million `(10^(6))` parts by mass in the solution `therefore"% bymass "=(15)/(10^(6))xx100=15xx10^(-4)` Taking 15 g chloroform in `10^(6)g` of the solution, mass of solvent `~=10^(6)g` Molar mass of `CHCl_(3)=12+1+3xx35.5="119.5 g mol"^(-1)` `therefore"Molality"=(15//119.5")/(10^(6))xx1000=1.25xx10^(-4)m.` |
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| 87612. |
A sample of dolomite contained 45% of CaCO_(3), 40% of MgCO_(3) and 15% clay. Calculate the mass of sulphuric acid of 30% strength required to react completely with 10 g of the sample. |
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| 87613. |
A sampleofdieaelhasthesamecharacteristics60mlmixtureofcentaneandalpha- methylnaphthalene mixedin 2:1 ratio(V/V) . Whatis thecetanenumberof thedieasel sample? |
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Answer» Solution :60 MLOF amixturecontains40 mLcetane and20mL of `alpha ` - METHYL naphthalene. Therefore, cetane number`=(40xx100)/(60)=66.6%` |
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| 87614. |
A sample of clay is found to have the formula Al_(2)O_(3).K_(2)O.6SiO_(2) . Calculate the percentage of alumina (Al_(2)O_(3)), potassium oxide (K_(2)O) and silica (SiO_(2)) in the sample. |
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Answer» `=102+94+360=556` `%" of "Al_(2)O_(3)=(102)/(556)xx100=18.35%, %" of "K_(2)O=(94)/(556)xx100=16.90%","%" of "SiO_(2)=(360)/(556)xx100=64` |
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| 87615. |
A sample of chloroform before being used as an anaesthetic is tested by: |
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Answer» `AgNO_3` solution |
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| 87616. |
A sample of charcoal weighing 6 g was brought into contact with a gas contained in a vessel of one litre capacity at 27^(@)C. The pressure of the gas was found to fall from 700 and 400 mm. Calculate the volume of the gas (reduced to STP) that is adsorbed per gram of the adsorbent under the condition of the experiment (density of charcoal sample is 1.5 g cm^(3)). |
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Answer» Solution :The adsorption is taking place in a closed vessel HENCE if pressure FALLS there is correspondingly increase in volume constant, excess of the volume of the gas would be adsorbed. `P_(1) V_(1) = P_(2) V_(2)` ltbr `V_(2) = P_(1) (V_(1))/(P_(2)) = 700 xx (1000)/(400) = 1750 mL.` Actual volume of the flask = 1000 - volume of charcoal `= 1000 - (6.00)/(1.50) = 996 mL.` Volume of the gas adsorbed = 1750 - 996 = 754 mL. Volume of the gas adsorbed per gram at STP `("USING " (P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))) = (125.67 xx 400 xx 273)/(300 xx 760) = 60.19 mL.` |
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| 87617. |
A sample of carbon from an ancient frame gives 7 counts of ""^(14)C per minute per gram of carbon. If freshly cut wood gives 15.3 counts of ""^(14)C per minute per gram, what is the age of the frame ? (Half-life period of ""^(14)C=5770 years) |
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| 87618. |
A sample of carbon dioxide that undergoes a transformation from solid to liquid and then to gas would undergo |
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Answer» a changein mass changeis stateofasubstancedoes notinvolvechangein masscomposition andphysicalpropertiesinstead ofinvolveschargeinvolumewhichalterdensity . |
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| 87619. |
A sample of calcium carbonate is 80% pure, 25 gm of this sample is treated with excess of HCl .How much volume of CO_(2) will be obtained from at 1 atm & 273 K ? |
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| 87620. |
A sample of CaCO_(3) has Ca=40%,C=12% and O=48% . If the law of constant proportions is true, then the mass of Ca in 5g of CaCO_(3) from another source will be: |
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Answer» 2.0g |
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| 87621. |
A sample of CaCO_3has Ca = 40%, C = 12% and O = 48%. If the law of constant proportion is true then the weight of calcium in 5 g of a sample of CaCO_3from another source will be |
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Answer» 0.20 G |
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| 87622. |
A sample of argon gas at l-atm pressure and 27^@Cexpands reversibly andadiabatically from 1.25 dm^3to 2.50 dm^3 . Calculate the enthalpy change in this process. C_(v,m)for argon is 12.48 JK^(-1) "mol"^(-1) |
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Answer» SOLUTION :T_1 = 300 K, cal `T_2 ` USING `T_2/T_1= (V_1/V_2)^(R//C_V)` `DELTA H = nC_p (T_2 - T_1) = n (C_v + R) (T_2 - T_1) ` -117.6 J |
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| 87623. |
A sample of Ca_(3)(PO_(4))_(2) contains 3.1 g phosphorous. The weight (in gm ) of Ca in the sample is |
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| 87624. |
A sample of an ideal gas is expanded from 1 dm^(3) to 3 in a reversible process for which P=KV^(3), with K=1//5 (atm//dm^(3)), what is work done by gas (L atm) |
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| 87625. |
A sample of ammonium phosphate, (NH_4)_3PO_4 contains 18 moles of hydrogen atoms. The number of moles of oxygen atoms in the sample is |
| Answer» ANSWER :`A` | |
| 87626. |
A sample of alloy contains a% copper (as CuAl_(2)) by mass. If % of CuAl_(2) in alloy is 7.40 then what is 'a'? |
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| 87627. |
A sample of AlF_3contains3.0 xx 10^24 F^(-)ions. The number of formula units of this sample is : |
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Answer» `2.0 xx 10^24 ` No. of `AlF_3` formula units` = (3.0 xx 10^24)/(3) = 1 xx 10^24 ` |
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| 87628. |
A sample of Al_(2)O_(3) dissolved in a molten fluoride bath is electrolysed using a current of 1.20 A. What is the rate of production of Al in g/hour ? The oxygen liberated at the positive carbon electrode reacts with the carbon to form CO_(2). What mass of CO_(2) is produced per hour ? |
| Answer» SOLUTION :0.403, 0.4924g/h | |
| 87629. |
A sample of alcohol at 200 torr and 70^(@)C is cooled at constant volume to 25^(@)C. What would exist at 25^(@)C? Vapour pressure of alcohol at 25^(@)C is 100 torr |
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Answer» Alcohol liquid and vapour at 100 torr |
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| 87630. |
A sample of air is saturated with benzene (v.p=100mm Hg) at298k, 750 mm Hg pressure . If is isothermally comressed to 1//3rd of initial volume, the final pressure is 410x. X is |
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| 87631. |
A sample of air is saturated with benzene (vapour pressure =100 mm Hg at 298 K) at 298 K, 750 mm Hg pressure.If it is isothermally compressed to one third of its initial volume, the final pressure of the system is |
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Answer» 2250 torr on COMPRESSING `P_f=650xx3` MM of Hg =1950 mm of Hg So, `P_T=(1950+100)=2050` mm of Hg |
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| 87632. |
A sample of air consisting mainly of N_2 and O_2 is heated at 2500K till the given equilibrium attains- N_2(g) + O_2 (g) hArr 2NO(g). If the value of equilibrium constant (K_c) at this temperature is 2.1xx10^(-3) and the percentage of number of moles of NO(g) in the above equilibrium mixture is 1.8 %, then what will be the mole fractions of N_2 and O_2 in the sample of air used'? |
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Answer» Solution :SUPPOSE, initial total NUMBER of moles of `N_2` and `O_2` in the sample of air = 1 mol. In the sample, if the initial number of moles of `N_2` is n, then the same for `O_2` is (1-n) mol. Let, the number of moles of `N_2` REACTS at equilibrium=`prop` Initial number of moles : `N_2+O_2 Number of moles at equilibrium: `(n-prop)(1-n-prop)2prop` Total number of moles at equilibrium `=n-prop+1-n-prop+2prop=1` so, mole percentage of NO`=(2prop)/(1_xx100=1.8` `:.prop=9xx10^(-3)` `K_c=([NO]^2)/([N_2][O_2])=(n_(NO)^2)/(n_(N_2)xxn_(O_2))` `=((2xx9xx10^(-3))^2)/((n-9xx10^(-3)(n-9xx10^(-3))=0.154` `:.` n=0.21 so, initial number of moles of `O_2`=0.21 and that of `N_2=1-0.21=0.79`. Hence, in the sample of air, mole FRACTION of `O_2`=0.21 and that of `N_2`=0.79 |
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| 87633. |
A sample of a salt has the percentage composition : Fe = 36.76 , S = 21.11 and O = 42.14 Calculate the empirical formula of the compound. (At. Mass Fe = 56, S = 32 and O = 16) |
| Answer» SOLUTION :`FeSO_(4)` | |
| 87634. |
A sample of a pure compound contains 2.04 grams of sodium, 2.65 xx 10^22atoms of carbon and 0.132 mole of oxygen atoms. Find the empirical formula of the compound |
| Answer» SOLUTION :`Na_2CO_3` | |
| 87635. |
A sample of a mixture of CaCl_(2) and NaCl weighing 4 . 22 g was treated to precipitateall Ca as CaCO_(3)whichwas then heated and quantitatively convertedto 0 . 959 g of CaO .Calculate thepercentage of CaCl_(2) in themixture.(Ca = 40 , O = 16 , C = 12 and Cl = 35 . 5) |
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Answer» Solution :We have, `{:(CaCl_(2)to CaCO_(3) to CAO),(" X g (say)0 . 959 g" ):}` Since Ca atomsare conserved, applying POAC for Ca atoms, moles of Ca in `CaCl_(2)` = molesof Ca in CaO `1 xx ` moles of `CaCl_(2) = 1 xx ` moles of CaO `(x)/( 111) = (0.959)/( 56) [{:(CaCl_(2) = 111),(" "CaO = 56 ):}]` x = 1.901 PERCENTAGE of `CaCl_(2) = 1 . 901 xx (100)/( 4 . 22 = 45 . 04%` |
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| 87636. |
A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of hydration is removed. The dried sample weighed 52 g. The formula of the hydrated salt is:(atomic mass, Ba = 137 am u, Cl = 35.5 am u) |
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Answer» `BaCl_(2).H_(2)O` `((52)/(208))`MOLE of `BaCl_(2)`with `((9)/(18))`moles of `H_(2)O "" :.`1 mole of `BaCl_(2)`with 2 moles of`H_(2)O` |
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| 87637. |
A sample of a gas expands from 200cm^(3) to 500cm^(3) against an average pressure of 750 torr while 1.5 J heat flows into the system. What is the change in energy of the system? (1 litre-atm=101.3J) |
| Answer» SOLUTION :ENERGY of the SYSTEM is DECREASED by 28.5 J, `DeltaE=-28.5J` | |
| 87638. |
A sample of a drug C_(21)_(23)O_(3)N (mol. Wt. =369) mixed with lactose C_(12)H_(22)O_(11) (mol.wt.=342) was analysed by osmotic pressure measurement to determine the amount of sugar present . If 100mL of solution containing 1.0g of the drug-sugar mixture has an osmotic pressure of 5.27mmHg at 25^(@)C , what is the per cent sugar present ? |
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Answer» SOLUTION :Let the amount of lactose be `xg`. `:.` moles of the SUGAR and drug MIXTURE `=(x)/(342)+((1-x))/(369)` Now, osmotic pressure `=cRT` `(527)/(760)=[{(x)/(342)+((1-x))/(369)}xx(1000)/(100)]xx0.0821xx298` `x=0.580g` `:.` PERCENTAGE of sugar `=58.0%` |
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| 87639. |
A sample of ""^238U (half-life = 4.5 xx 10^9 yr ) ore is found to contain 23.8 g of ""^238U and 20.6 g of""^206Pb . Calculate the age of the ore. |
| Answer» SOLUTION :`4.489 XX 10^9 `YEARS | |
| 87640. |
A sample of ._(19)^(40)K contains invariably ._(18)^(40)Ar. This is because ._(19)^(40)K has tendency to undergo: |
| Answer» Answer :b | |
| 87641. |
A sample of 16gr of charcoal was brought into contact with CH_4 gas contained in a vessel of llitre at 27^@C the pressure of gas was found to fall from 760 to 608 torr. The density of charcoal sample is 1.6gr//cm^3. What is the volume of the CH_4 gas adsorbed per gram of the adsorbent at 608 torr and 27^@C |
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Answer» 125ml/gr volume occupied by gas = volume of vessel-volume occupied by charcoal `= 1000- 16/1.6 = 990 ml`. DIFFERENCE of volume is due to adsoprption of gas by characoal `:.` volume of gas adsorbed by charcoal = `1250 - 990 = 260 ml ` volume of the gas adsorbed per GRAM of characoal = `260/16 = 16.25 ml//gr^(+)` at 608 torr and `27^@C` |
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| 87642. |
A sample of ._(38)^(90)Sr has an activity of 0.5 mCi. What is its specific activity? (t_(1//2) of ._(38)^(90)Sr = 19.9 years) |
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Answer» SOLUTION :Rate of disintegrations `= LAMBDA xx` No. atoms So, No of atoms `= (0.5 xx 3.7 xx 10^(7))/(0.693) xx 19.9 xx 365 xx 24 xx 60 xx 60` `= 1.675 xx 10^(16)` Mass = `(90 xx 1.675 xx 10^(13))/(6.023 xx 10^(23)) = 2.50 xx 10^(-6) g` Specific activity `= (0.5 xx 3.7 xx 10^(7))/(2.5 xx 10^(-6)) = 7.4 xx 10^(12) "DIS" g^(-1) s^(-1)` |
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| 87643. |
A sample of ""^14CO_2was to be mixed with ordinary CO_2for biological tracerexperiment. In order that 10 cc (NTP) of the diluted gas should have 104 disintegrations per minute, how many microcuries of radioactive carbon are needed to prepare 60 litres of the diluted gas? |
| Answer» SOLUTION :`27 MU CI` | |
| 87644. |
A sample of .^(14)CO_(2) was to be mixed with ordinary CO_(2) for a biological tracer experiment. In order that 10 "cm"^(3)of diluted gas should have 10^(4) dis/min,what activity ("in"mu Ci) of radioactive carbon is needed to prepare 60L of diluted gas at 1 atm and 273 K? [1 Ci=3.7 xx10^(10) "dps"] |
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Answer» `270mu CI` |
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| 87645. |
A sample of .^(14)CO_(2) was to be mixed with ordinary .Co_(2) for a biological tracer experiment. In order that 10^(3) cm^(3) of the diluted gas at NTP should have 10^(4) dis/min, how many muCi Of radiocarbon-14 are needed to prepare 60 L ofthe diluted gas ? |
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Answer» SOLUTION :`10 cm^(3)` of the diluted gas at NTP `10^(4)` dis/min `(10^(4))/(60)` DPSL `:.` 60 L (60, 000 `cm^(3)`) of the dilute gas at NTP `(10^(4) xx 60,000)/(60 xx 10)` dps Thus, no of `MU Ci` of `.^(14)CO_(2)` needed `= (10^(4) xx 60,000)/(60 xx 10 xx 3.7 xx 10^(4)) (1 mu Ci = 3.7 xx 10^(4) dps)` `= 27.03 mu Ci` |
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| 87646. |
A sample of 10gm of H_(2) reacts with sufficient amount of oxygen to form 106gm of H_(2)O(l) and H_(2)O_(2)(l). Calculate mass % of H_(2)O. |
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Answer» 0.679 B mole `H_(2)O_(2)` `axx18+bxx34=106` . . . (i) `POAC` on H: `2xxn_(H_(2))=2xxn_(H_(2)O)+2xxn_(H_(2)O_(2))` `2XX(10)/(2)=10=2a+2b` `a+b=5` . . . .(ii) `18a+18b=90`. . . . . . ..(ii) `18a+18b=106` . . . . . . .(i) 16b=16 b=1 a=4 Mass of `H_(2)O=axx18` `=4xx19=72gm` Mass `% H_(2)O=(72)/(106)xx100=67.9%` |
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| 87647. |
A sample of 100 mL of a solution of a weak monoprotic acid of unknown concentration is titrated with 0.500 M NaOH to give the titration curve shown.All of the statements are correct except: |
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Answer» PHENOLPHTHALEIN would be a suitable indicator for this TITRATION |
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| 87648. |
A sampleof 0.15 g of the compound [Pt(NH_(3))_(x)Br_(y)]^(z+) .z Br^(-) , ignited and heated to decomposition produced 0.0502g of Pt. A second0.15 g sample was dissolved in water and titrated rapidly with 0.01 " M " AgNO_(3) solution . 51.50 mL was requiredto precipitate all theionicbromie . A third 0.15 - g sample was heatedfor two hours on a steam both in a solution to which0.2 moleofAgNO_(3) has been added . this precipitated all the bromide ( not just the free ionic Br^(-) ) as AgBr. The weightof the precipitatedthusproduced was 0.20 g . Find x , y and z(Pt = 195 , ag = 108 , Br = 80 , N = 14 and H = 1) |
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Answer» Solution :`[Pt(NH_(3))_(x)Br_(y)]^(2+) .z BR^(-) overset(Delta) to Pt` Applying PoAC for Pt atoms , ` 1 xx ` mole of the COMPOUND = `1XX ` mole of Pt `(0.15)/M = (0.0502)/195""…(1)` where `M = 195 +17x +80 y + 80 z "" ..(2)` For `0.15`G of the secondsample containing z Br atoms molecule( only the ionic bromide ) , one moleculeof the compound shall combine with z molecules of `AgNO_(3)`to give z molecules of AgBr . Applying mole ratio method to reactants `Z xx ` mole of the compound = `1 xx` mole of `AgNO_(3)` `z = (0.15)/M = (0.01 xx 51.5)/1000 ""...(3)` For the 0.15 g of the thirdsample , all bromine atoms `(y+z)`in the compound combine with `AgNO_(3)` to give `(y+z)` molecules of AgBr , Applying mol ratio method , `(y+z) xx` mole of the compound = `1 xx ` mole of AgBr ` (y+z) xx (0.15)/M = 1 xx (0.20)/188 "" ...(4)` Solving equations (1) , (2) ,(3) and (4) we get , x = 4 , y = 2 and z = 2 . |
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| 87649. |
A sample contains two radioactive substances A and B in the ratio of 4:1. If their half lives are 12 and 16 has respectively them after two days. What will be the ratio of A and B ? |
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Answer» `1:1` |
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| 87650. |
A sample f O_(2) gas is collected over water at 23^(@)C at a barometric pressure of 751 mm Hg ( vapour pressure of water at 23^(@)C is 21 mm Hg). The partial pressure of O_(2) gas in the sample collected is |
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Answer» 21 MM HG `= (730)/(760)=0.96 atm` |
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