every option with multiple choice want ans of every option and it is a multiple choice match column Share with your friends Share 0 Vartika Jain answered this Dear Student, (A)100 g of compound contains 52.17g of C, 13.04 g of H and 34.78 g of O. Number of moles of C=52.1712=4.34 molNumber of moles of H=13.041=13.04 molNumber of moles of O=34.7816=2.17 molDividing each by the lowest, seeking the smallest whole number ratio which is 2.14 in this caseC=4.342.17=2H=13.042.17=6O=2.172.17=1Therefore, empirical formula =C2H6OEmpirical formula mass=2×12+6×1+16=46 Number of Empirical units=Molar massEmpirical formula mass=4646=1Hence molecular formula=(C2H6O)1=C2H6OHence, empirical formula=molecular formulaTherefore, correct option is (q) (B)Mass of organic compound – 0.30gThe compound contains C , H & O elements.Mass of CO2 gas produced after combustion =0.44gMass of H2O produced after combustion = 0.18gCalculation of mass of carbon:C + O2 → CO2 12g 44g44g of CO2 contains 12g of carbon0.44g of CO2 contains 0.44×12 44 = 0.12gCalculation of mass of Hydrogen2H2 + O2 → 2H2O4g 36g36g of water contains 4g of hydrogen0.18g of water contains 0.18436 = 0.02g% of C = Mass of carbon Mass of compound × 100= 0.120.30 × 100= 40%% of H= 0.020.30 × 100 = 6.6%% of oxygen = 100 -(%C +% H)= 100 – 46.66= 53.34% Element % composition % / Atomic weight Simple ratio Carbon 40% 40/12 = 3.33 3.33/3.33 = 1 Hydrogen 6.66% 6.66/1 = 6.66 6.66/3.33 = 2 Oxygen 53.34% 53.34/16 =3.33 3.33/3.33 = 1Number of carbon atoms present in the given compound is 1Number of hydrogen atoms present in the given compound is 2 Number of oxygen atoms present in the given compound is 1 The empirical formula of given compound is CH2OBut here it isgiven than 1 molecule contains 2 O atomTherefore, a shoud be multiplied by 2i.e. Molecular formula = C2H4O2Molecular mass = 2×12+4×1+16×2=60 gNow, 1 molecule of compound contains 4 H atomand 1 mol contains NA moleculesTherefore, 1 mol of compound contains 4NA H atomsHence, correct answer is (p) (D)Molecular mass=2×V.D.or, Molecular mass=2×46=92 gWeight of carbon =10.5 gWeight of hydrogen =1 gNumber of moles of C=10.512=0.875 molnumber of moles of H=11=1 molSimple ratio of C=0.8750.875=1Simple ratio of H=10.875=1.14Whole number of C=7Whole number of H=8Hence, empirical formula=C7H8Empirical mass=12×7+8×1=92 gNumber of empirical units=molecular massempirical formula mass=9292=1Hence,molecular formula=empirical formula=C7H8C7H8 + 9O2 → 7CO2 + 4H2O1 mol of hydrocarbon on combustion gives 7 moles of CO2 and 4 moles of H2OTherefore, moles of CO2 is greater than moles of H2OHence, the correct options are (q) and (r) Similarly, answer can be calculated for C also 0 View Full Answer