(b) 1 mole C2H6C_{2}H_{6}C2​H6​ contains six moles of H- atoms. = Number  of  moles  of  soluteVolume  of  solution  in  Litres\frac{Number\;of\;moles\;of\;solute}{Volume\;of\;solution\;in\;Litres}VolumeofsolutioninLitresNumberofmolesofsolute​, = Mass  of  sugarMolar  mass  of  sugar2  L\frac{\frac{Mass\;of\;sugar}{Molar\;mass\;of\;sugar}}{2\;L}2LMolarmassofsugarMassofsugar​​, = 20  g[(12  ×  12)  +  (1  ×  22)  +  (11  ×  16)]g]2  L\frac{\frac{20\;g}{[(12\;\times \;12)\;+\;(1\;\times \;22)\;+\;(11\;\times \;16)]g]}}{2\;L}2L[(12×12)+(1×22)+(11×16)]g]20g​​, = 20  g342  g2  L\frac{\frac{20\;g}{342\;g}}{2\;L}2L342g20g​​, = 0.0585  mol2  L\frac{0.0585\;mol}{2\;L}2L0.0585mol​, Therefore, Molar concentration = 0.02925 molL−1L^{-1}L−1. 2 volumes of dihydrogen react with 1 volume of dioxygen to produce two volumes of vapour. Similarly 2.5 moles of X reacts with 2 moles of Y, so 2.5 mole of X is unused. = 1  ×  1031 \; \times \;10^{ 3 }1×103 – 428.6 g. Q25. 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Submit your article Opens in new window Information and templates for authors Search this journal. Numerical problems in calculating mass percent and concentration. Return within 21 days of the order for any reason. Thus, 100 g of HNO3 contains 69 g of HNO3 by mass. Calculate the number of atoms in each of the following, 1 mole of Ar = 6.023  ×  10236.023 \; \times \; 10^{ 23 }6.023×1023 atoms of Ar, Therefore, 52 mol of Ar = 52 × 6.023  ×  10236.023 \; \times \; 10^{ 23 }6.023×1023 atoms of Ar, = 3.131  ×  10253.131 \; \times \; 10^{ 25 }3.131×1025 atoms of Ar, 1 u of He = 14\frac{ 1 }{ 4 }41​ atom of He, 52 u of He = 524\frac{ 52 }{ 4 }452​ atom of He, 4 g of He = 6.023  ×  10236.023 \; \times \; 10^{ 23 }6.023×1023 atoms of He, 52 g of He = 6.023  ×  1023  ×  524\frac{ 6.023 \; \times \; 10^{ 23 } \; \times \;52 }{ 4 }46.023×1023×52​ atoms of He, = 7.8286  ×  10247.8286 \; \times \; 10^{ 24 }7.8286×1024 atoms of He.