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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.
| 24451. |
The ratio of the amount of two element X and Y at radioactive equilibrium is 1 : 2 xx 10^(-6). If the half-life period of element Y is 4.9 xx 10^(-4) days, then the half-life period of element X will be |
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Answer» `4.8 xx 10^(-3)` days |
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| 24452. |
The ratio of speeds of diffusion of two gases A and B is 1:4. If the mass ratio of A to B present in the given mixture is 2:1, then which of the following is the ratio mole-fraction of A to B? |
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Answer» `2:3` `(1)/(4)=(m_(A)//M_(A))/(m_(B)//M_(B))sqrt((M_(B))/(M_(A)))` `(1)/(4)=(m_(A))/(m_(B))((M_(B))/(M_(A)))^(3//2)` `(1)/(4)=(2)/(1)((M_(B))/(M_(A)))^(3//2)` `((M_(B))/(M_(A)))(8)^(2//3)=(1)/(4)` `(n_(A))/(n_(B))=(m_(A))/(m_(B))XX(M_(B))/(M_(A))=(2)/(1)xx(1)/(4)` =1:2 |
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| 24453. |
The ratio of specific heat at constant pressure to specific heat at constant volume (i.e. C_p//C_V) for noble gases is |
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Answer» 1.66 |
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| 24454. |
The ratio of specific charge (e/m) of an electron to that of a hydrogen ion is: |
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Answer» 1:1 |
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| 24455. |
The ratio of specific charge (e//m) of a proton and that of an alpha-particle is: |
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Answer» |
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| 24456. |
The ratio of sigma and pi bonds present in XeO_4 molecule is |
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Answer» `2 : 3` |
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| 24457. |
The ratio of sigma to pi bonds is benzene is : |
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Answer» 2 |
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| 24458. |
The ratio of sigma and pi-bonds in benzene is |
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Answer» 2 |
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| 24459. |
The ratio of r.m.s. velocity and average velocity of a gas molecule at a particular temperature is : |
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Answer» `1.086 :1` |
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| 24460. |
The ratio of relative rates of isopropyl bromide and ethyl bromide in S_N1 reaction is |
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Answer» `11: 1` |
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| 24461. |
The ratio of rates of diffusion of hydrogen chloride and ammonia gases is |
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Answer» `1: 1.46` |
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| 24462. |
The ratio of radius of 4th orbit of hydrogen and 3rd orbit of Li^(2+) is |
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Answer» `256:9` `r_(4)(H)=(0.059xx16)/1` `r_(3)(Li^(2+))=(0.059xx9)/3, (r_(4)(H))/(r_(3)(Li^(2+)))=16/3` |
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| 24463. |
The ratio of rate constants of a reaction at 300K and 291K is 2. Calculate the energy of activation. ("Given "R = "8.314JK"^(-1)" mol"^(-1)). |
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Answer» SOLUTION :To find activation ENERGY for the reaction Given `(K_(2))/(K_(1))=2, R=8.314, T_(1)=291K, T_(2)=300K` `LOG"" (K_(2))/(K_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `log2=(E_(a))/(2.303xx8.314)[(300-291K)/(300xx291K)]` `E_(a)=(2.303xx8.314xxlog2xx300xx291)/(9)=55.9kJ` |
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| 24464. |
The ratio of rates of diffusion of CO_2 and SO_2 at the same P and T is : |
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Answer» 4:sqrt11 |
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| 24465. |
The ratio of rates of diffusion of CO_2 and SO_2 at the same temperature and pressure will be |
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Answer» `4 : sqrt 11` |
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| 24466. |
The ratio of radii of the first three Bohr orbits of H-atom is : |
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Answer» `1:2:3` |
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| 24467. |
The ratio of partial pressure of a gaseous component to the total vapour pressure of the mixture is equal to : |
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Answer» MASS of the component |
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| 24468. |
The ratio of p_(pi)-d_(pi) bonds is SO_(2) and SO_(3) molecules |
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Answer» `1:1` |
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| 24469. |
The ratio of oxygen atom havingoxidation no. inS_(2)O_(B)^(2-)is : |
Answer» 
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| 24470. |
The ratio of number of single and double P—Q bonds in P_(4)O_(10)is/are ____ |
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Answer» <P>  `{:(P-O " BONDS" =4),(P-O" bonds"),("Single P - O"),(BAR"Single P - O"=12/4=3):} 12/4=3` |
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| 24471. |
The ratio of number of oxygen atoms (O) in 16.0 g ozone (O_(3)), 28.0 g carbon monoxide (CO) and 16.0 g oxygen (O_(2)) is :- (Atomic mass : C = 12, O = 16 and Avogadro's constant N_(A) = 6.0 xx 10^(23) mol^(-1)) |
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Answer» `3 : 1 : 1` |
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| 24472. |
The ratio of number of d pi - p pi bonds in XeO_(4) and SO_(3) is. |
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Answer» |
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| 24473. |
The ratio of nucleons in O^16 and O^18 is: |
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Answer» `8/9` |
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| 24474. |
The ratio of most probable velocity, average velocity and root mran square velocity is |
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Answer» `SQRT2 : SQRT(8/pi) : SQRT3 ` |
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| 24475. |
The ratio of masses of oxygen and nitrogen in as particular gaseous mixture is 1 : 4. The ratio of number of their molecule is : (Atomic mass Ag = 108, Br = 80) |
| Answer» Answer :D | |
| 24476. |
The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1:4. The ratio of the number of their molecules is |
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Answer» `3:16` Then, mass of `N_(2)` in the mixture = 4 x g No. of moles of `O_(2)=(x)/(32)` No. of moles of `N_(2)=(4X)/(28)` As molecules are in the RATIO of their moles, ratio of their molecules is `O_(2):N_(2)=(x)/(32):(4x)/(28)=(1)/(32):(1)/(7)=7:32` |
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| 24477. |
The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is |
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Answer» `1 : 4` `= ((m_(O_(2)))/(m_(N_(2)))) (28)/(32) = (1)/(4) xx (28)/(32)` `= (7)/(32)` |
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| 24478. |
The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1:4. the ratio of number of their molecules is: |
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Answer» `1:8` `=n_(O_(2)):n_(H_(2))` `=((w)/(m))_(O_(2)):((w)/(m))_(N_(2))` `=(1)/(32):(4)/(28)=7:32` |
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| 24479. |
The ration of mass per cent of C and H of an organic compound (C_(x)H_(y)O_(z)) "is"6:1. If one molecule of the above compound (C_(x)H_(Y)O_(z)) contains half as much oxygen as required to burn one molecule of compound C_(x)H_(Y) compleltely to CO_(2) and H_(2)O. The empirial formula of compound C_(x)H_(y)O_(z) is: |
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Answer» `C_(3)H_(6)O_(3)` |
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| 24480. |
The ratio of mass percent of C and H of an organic compound (C_(X)H_(Y)O_(Z)) is 6:1. If one molecule of the above compound (C_(X)H_(Y)O_(Z)) contains half as much oxygen as required to burn one molecule of compound C_(X)H_(Y) completely to CO_(2) and H_(2)O, the empirical formula of the compound C_(X)H_(Y)O_(Z) is |
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Answer» `C_(3)H_(6)O_(3)` As `C:H=6:1`, we have  `thereforeX=1, Y=2` Equation for combustion of `C_(X)H_(Y)` is |
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| 24481. |
The ratio of mass percent of C and H of an organic compound (C_(X)H_(Y)O_(Z)) is 6 : 1 If one molecule of the above compound (C_(X)H_(Y)O_(Z))contains half as much oxygen as required to burn one molecule of compound C_(X)H_(Y) completely to CO_(2) and H_(2)O. The empirical formula of compound C_(X)H_(Y)O_(Z) is |
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Answer» `C_(2)H_(4)O` |
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| 24482. |
The ratio of mass per cent of C and H of an organic compound (C_(x)H_(y)O_(z)) is 6:1. If one molecule of the above compound contains half as much oxygen as required to burn one molecule of the compound C_(x)H_(y) completely to CO_(2) and H_(2)OThe empirical formula of the compound C_(x)H_(y)O_(z) is |
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Answer» `C_(2)H_(4)O_(3)` `overset("1 mole")(C_(x)H_(y)) + overset("z mole")(O_(2)) rarr CO_(2) + H_(2)O` , applying POAC for C, H and O, we GET, `z=x + (y)/(4)`, if x=1 and y= 2` therefore z=1.5` i.e., `C_(2)H_(4)O_(3)` should be the EMPIRICAL formula |
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| 24483. |
The ratio of mass of 1 mole of sodium and 10^23atoms of sodium is : |
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Answer» 6.02 ` = (23)/((6.02 xx 10^23) xx 10^23) = 6.02` |
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| 24484. |
The ratio of (Lambda_(m)/Lambda_(e)) for Ca_(3) (PO_(4))_(2) will be equal to : |
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Answer» |
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| 24485. |
The ratio of ((Lambda_(m))/(Lambda_(c))) for Ca_(3)(PO_4)_2 will be equal to ........... |
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Answer» |
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| 24486. |
The ratio of K_(p)//K_(c ) for the reaction :N_(2)(g)+O_(2)(g) hArr 2NO(g)is |
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Answer» 4 |
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| 24487. |
The ratio of kinetic energy and potential energy of an electron in any orbit is equal to: |
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Answer» Zero |
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| 24489. |
The ratio of initial activities of two samples of redionuclides A and B are ((A_(A))/(B_(B))) = 1/(16). If t_(1//2) (A) "and" t_(1//2) (B) "are" 30min and 7.5 minactivity of both samples will be same ? |
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Answer» |
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| 24490. |
The ratio of heats liberated at 298 K from the combustion of one kg of coke and by burning water gas obtained from 1kg of coke is (assume coke to be 100% carbon). (Given: Enthalpies of combustion of CO_(2),CO and H_(2) are 393.5 kJ, 285 kJ, 285 kJ respectively all at 298K) |
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Answer» `0.79:1` For the combustion of 1 kg of coke `C+O_(2) to CO_(2)""DeltaH=393.5kJ` `implies`HEAT liberated from 1 mole of coke=393.5 kJ `therefore`Heat liberated from 83.33 mole of coke `H_(1)=(393.5xx83.33)kJ` Also, for the BURNING of water gas `C+H_(2)O to underset("Water gas")ubrace(CO+H_(2))` `CO+H_(2)+O_(2)toCO_(2)+H_(2)O,DeltaH=285+285=570kJ` `therefore`Ratio of heat liberated from burning of water gas obtained from 1 mole of coke `H_(2)=(570xx83.33)kJ` `therefore`REQUIRED ratio `=H_(1):H_(2)=393.5:570=0.69:1` |
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| 24491. |
The ratio of heats liberated at 298 K from the combustion of one kg of coke and by burning watergas obtained from kg of coke is (Assume coke to be 100% carbon). (Given enthalpies of combustion of CO_2, CO and H_2 as 393.5 kJ, 285 kJ, 285 kJ respectively at 298 K). |
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Answer» `0.79 : 1` `C + O_(2) rarr CO_(2)` Heat liberated by BURNING 83.33 mol of coke `= 83.33 xx 393.5 kJ` `ubrace(CO + H_(2))_("water GAS") + O_(2) rarr CO_(2) + H_(2)O` Heat liberated by burning 1 mol of water gas `= 285 + 285 = 570 kJ`. Heat liberated by burning water gas obtained from 1 kg of coke `= 83.33 xx 570 kJ` Ratio of heat liberated Coke : water gas `83.33 xx 393.5 : 83.33 xx 570` `393.5 : 570` `0.69 : 1` |
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| 24492. |
The ratio of gamma = (C_(p) //C_(v))for inert gases is : |
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Answer» 1.33 |
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| 24493. |
The ratio of Fe^(3+) and Fe^(2+) ions in Fe_(0.9)9S_(1.0) is |
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Answer» 0.28 |
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| 24494. |
The ratio of everage speed, most probable speed and root mean square speed of a gas sample will be |
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Answer» `sqrtfrac{4}{PI}:SQRT3:SQRT2` |
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| 24495. |
The ratio of enthalpy of vaporisation and the normal boilng point of a liquid approximately equals to (i)_____. The above statement follows the (ii)______. Choose the correct word to complete the above statement. |
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Answer» (i) 88 J `mol^(-1)K^(-1)`, (ii) capacity rule the above statement follows the trouton's rule. |
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| 24496. |
The ratio of energy of photons of ddote= 2000 dotA" to that of " ddote= 4000 dotA is |
| Answer» ANSWER :C | |
| 24497. |
The ratio of energy of a photon of 2000Å wavelength radiation to that of 4000Å radiation is |
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Answer» `1//4` |
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| 24498. |
The ratio of energy of a photon of 2000Å wavelengt radiation to that 4000Å radiation is |
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Answer» 43834 |
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| 24499. |
The ratio of electron , proton and neutron in tritium is : |
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Answer» 1 : 1 : 1 |
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| 24500. |
The ratio of difference in energy between the first and second Bohr orbits to that of second and third Bohr orbits is |
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Answer» `1//2` |
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