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Some Basic Concepts of Chemistry Test - 47

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Some Basic Concepts of Chemistry Test - 47
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  • Question 1
    1 / -0
    A mixture contains equimolar quantities of carbonates of two bivalent metals. One metal is present to the extent of 13.5%13.5\% by weight in the mixture and 2.582.58g of the mixture on heating leaves a residue of 1.351.35g. What percentage by weight of the other metal is there?
    Solution
    Let equimolar metal carbonates have mole be nn
    Let Metal be M1M_1 and M2M_2
    As M1CO3Δ M1O+CO2(g){ M }_{ 1 }{ CO }_{ 3 }\underrightarrow { \Delta  } { M }_{ 1 }O+{ CO }_{ { 2 }_{ (g) } }\uparrow
         M2CO3Δ M2O+CO2(g){ M }_{ 2 }{ CO }_{ 3 }\underrightarrow { \Delta  } { M }_{ 2 }O+{ CO }_{ { 2 }_{ (g) } }\uparrow
    Let molecular wt. of M1w1{ M }_{ 1 }\rightarrow { w }_{ 1 }
          molecular wt. of M2w2{ M }_{ 2 }\rightarrow { w }_{ 2 }
    for % composition of metal M1=nw1n(w1+60)+n(w2+60) =0.135(1){ M }_{ 1 }=\cfrac { n{ w }_{ 1 } }{ n\left( { w }_{ 1 }+60 \right) +n\left( { w }_{ 2 }+60 \right)  } =0.135\quad -(1)
    \Rightarrow we have n(w1+60)+n(w2+60)=2.58gm(2)n\left( { w }_{ 1 }+60 \right) +n\left( { w }_{ 2 }+60 \right) =2.58gm\quad -(2)
           CO2CO_2 librated=2.581.35=1.23gm=2.58-1.35=1.23gm
      \therefore wt. of CO2CO_2 in mixture=n(44)+n(44)=n(44)+n(44)
                                              =88ngm=88n\quad gm
          88ngm=1.23gm88n\quad gm=1.23gm
                         n=1.2388=0.01397n=\cfrac {1.23}{88} =0.01397
    Put n=0.01397n=0.01397 mol in (1)(1) we get
         0.01397w12.58=0.135\cfrac {0.01397 w_1}{2.58} =0.135
                     [w1=24.93gm][w_1 =24.93gm]
    From (2)(2)
      n(w2+w1+120)=2.58n(w_2 +w_1 +120)=2.58
    0.01397(24.93+w2+120)=2.580.01397(24.93+ w_2 +120)=2.58
       2.0246+0.01397w2=2.582.0246+0.01397 w_2 =2.58
                                     [w2=39.75gm][w_2 =39.75gm]
    % composition of M2=nw22.58=0.01397×39.752.58=0.215M_2 =\cfrac { n{ w }_{ 2 } }{ 2.58 } =\cfrac { 0.01397\times 39.75 }{ 2.58 } =0.215
                                                                                            =21.5=21.5%
  • Question 2
    1 / -0
    A precipitate of AgClAgCl and AgBrAgBr has the mass 0.40660.4066 g. On heating a current of chlorine, the AgBrAgBr is converted to AgClAgCl and the mixture loses 0.07250.0725 in mass. The %% of mass in the original mixture is (as the nearest integer) :
    Solution
    Option B is correct

  • Question 3
    1 / -0
    The leaves of a green bean plant were exposed to carbon dioxide containing the radioactive isotope of carbon 14C^{14}C for few hours in sunlight. A section of the plant's stem was then immediately obtained and placed on an X-ray film that blackens where exposed to radioactivity. The following diagrams show the cross-section of the stem, and the resulting autoradiograph. Why is structure PP shown radioactive?

  • Question 4
    1 / -0
    Decreasing order of atomic weight is correct of the elements given below?
    Solution
    Atomic weight of Iron  56\approx 56 u
    Atomic weight of Cobalt 59\approx 59 u
    Atomic weight of Nickel 58.7\approx 58.7 u
    So, the order of atomic weights is Co>Ni>FeCo>Ni>Fe.
    Option CC is correct.
  • Question 5
    1 / -0
    The molecular weights of O2O_2 and N2N_2 are 3232 and 2828 respectively. At 15015^0C, the pressure of 11gm O2O_2 will be the same as that of 11 gm N2N_2 in the same bottle at the temperature:
    Solution
    Temperature of O2O_{2} =273+15=298K=273+15=298 K

    Temperature of N2N_{2} =x=x

    As nn \propto 1T\dfrac 1T, as pressure is constant.

    n1T1=n2T2(132 )(298)=(128 )(x)x=2832×298=260.75K\Longrightarrow { n }_{ 1 }{ T }_{ 1 }={ n }_{ 2 }{ T }_{ 2 }\Longrightarrow \left( \cfrac { 1 }{ 32 }  \right) \left( 298 \right) =\left( \cfrac { 1 }{ 28 }  \right) \left( x \right) \\ \Longrightarrow x=\cfrac { 28 }{ 32 } \times 298=260.75K

    Temperature in oC=260.75273=13oC^oC=260.75-273=-13 ^oC

    Hence, option BB is correct.
  • Question 6
    1 / -0
    Calculate the total number of oxygen atoms present in 1.80g1.80g of glucose (C6H12O6)({C}_{6}{H}_{12}{O}_{6}).
    Solution
    GlucoseC6H12O6Glucose\longrightarrow C_6H_{12}O_6
    $$Number\  of\  moles=\dfrac{mass}{molecular\ mass}$$
    =1.8180=\dfrac{1.8}{180}
    =0.01 mol=0.01\ mol
    Now, there are 6 atoms of oxygen
    Total number of oxygen atoms==number of atoms of oxygen×\timesnumber of moles×\timesAvagadro's constant
                                                       =6×0.01×6.023×1023=6\times0.01\times6.023\times10^{23}
                                                       =36.138×1021=36.138\times10^{21}
  • Question 7
    1 / -0
    What will be the mass percentage of a resulting solution prepared by mixing 15%15\% (w/w) 500 g aqueous solution of the area with 25%25\% (w/w) 400g aqueous solution of it?
    Solution
    Given (w/w)%({w/w})\% of 1st1^{st} solution =15=15
        15=500Weightofsolution(x1)×100\implies 15=\cfrac{500}{Weight\quad of\quad solution(x_1)}\times 100
        x1=(500)(100)15\implies x_1=\cfrac{(500)(100)}{15}
        x1=1043\implies x_1=\cfrac{10^{4}}{3}
    Given (w/w)%({w/w})\% of 2nd2^{nd} solution =25=25
        25=400Weightofsolution(x2)×100\implies 25=\cfrac{400}{Weight\quad of\quad solution(x_2)}\times 100
        x2=(400)(100)25\implies x_2=\cfrac{(400)(100)}{25}
        x2=1600\implies x_2=1600
    Total (w/w)%=[500+4001043+1600]×100=90033.3+16=90049.318({w/w})\%=\left[\cfrac{500+400}{\cfrac{10^4}{3}+1600}\right]\times 100=\cfrac{900}{33.3+16}=\cfrac{900}{49.3}\simeq 18
    \therefore Weight by weight percentage of resulting solution is 18%18\%
  • Question 8
    1 / -0
    The critical volume of a gas is 0.036 lit.mol1lit. mol^{-1}. The radius of the molecule will be (in cm):
    (Avogadro Number = 6×10236 \times 10^{23})
    Solution
    Critical volume of gas=3b [for 1 molecule]
    For 1 mole, critical volume=0.036 liters
    0.036×103cm3=3(43πr3)×NAr=(3624π ×1023) 13 =(38π ×1023) 13 \Longrightarrow 0.036\times { 10 }^{ 3 }{ cm }^{ 3 }=3\left( \cfrac { 4 }{ 3 } \pi { r }^{ 3 } \right) \times { N }_{ A } \\ \Longrightarrow r={ \left( \cfrac { 36 }{ 24\pi  } \times 10^{ -23 } \right)  }^{ \cfrac { 1 }{ 3 }  }={ \left( \cfrac { 3 }{ 8\pi  } \times 10^{ -23 } \right)  }^{ \cfrac { 1 }{ 3 }  }
  • Question 9
    1 / -0
    What do carbon credits signify?
    Solution
    A carbon credit is a generic term for any tradable certificate or permit representing the right to emit one tonne of carbon dioxide or the mass of another greenhouse gas with a carbon dioxide equivalent, equivalent to one tonne of carbon dioxide.
  • Question 10
    1 / -0
    What mass of propene is formed from 34.0 g of iodopropane on heating with ethanolic KOHKOH, if the yield is 36 %?
    Solution
    CH3CH2CH2I+KOHCH3CH=CH2+KI+H2OCH_3CH_2CH_2I+KOH \longrightarrow CH_3-CH=CH_2+KI+H_2O
    Moles of Iodopropane =34170=0.2=\cfrac{34}{170}=0.2 molesmoles
    11 molemole of iodopropane gives 11 molemole propene but yield is 36%36\%.
    So, 11 molemole of iodopropane will give 0.360.36 molemole of propene.
    0.20.2 molemole of iodopropane will give 0.2×0.36=0.0720.2\times 0.36=0.072 molemole of propene.
    Mass of propene =42×0.072=3.024=42\times 0.072=3.024 gramgram
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