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Chemistry Question on Percentage Composition

The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
Mass of dinitrogen Mass of dioxygen
(i) 14 g 16 g
(ii) 14 g 32 g
(iii) 28 g 32 g
(iv) 28 g 80 g

  1. Which law of chemical combination is obeyed by the above experimental data? Give its statement.
  2. Fill in the blanks in the following conversions:
    (i)1 km = ...................... mm = ...................... pm
    (ii) 1 mg = ...................... kg = ...................... ng
    (iii) 1 mL = ...................... L = ...................... dm3
Answer

(a) If we fix the mass of dinitrogen at 28  g28 \;g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 32  g32 \;g, 64  g64 \;g, 32  g32 \;g, and 80  g80 \;g. The masses of dioxygen bear a whole number ratio of 1:2:2:51:2:2:5.
Hence, the given experimental data obeys the law of multiple proportions.
The law states that if two elements combine to form more than one compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers.


(b) (i) 1  km=1\; km = 1  km×1000m1km×100cm1m×10mm1cm1\; km × \frac{1000 m }{ 1 km} × \frac{100 cm }{ 1 m} × \frac{10 mm }{ 1 cm}
1km=106mm1 km = 10 ^6 mm
1  km=1\; km = 1km×1000m1km×1pm1012m1 km × \frac{1000 m }{ 1 km} ×\frac{ 1 pm }{ 10 ^{-12} m}
1  km=1015pm1\; km = 10 ^{15} pm

Hence, 1  km1\; km = 106mm10^ 6 mm = 1015  pm10 ^15 \;pm

(ii) 1  mg=1  mg×1\; mg = 1\; mg × 1g1000mg×1kg1000g\frac{1\, g }{ 1000 \,mg} × \frac{1 \,kg }{ 1000 \,g}
1  mg=\Rightarrow 1\; mg = 106  kg10 ^{-6}\; kg
1  mg=1  mg×1g1000mg×1ng109g1\; mg = 1\; mg × \frac{1 g }{ 1000 mg} × \frac{1 ng }{ 10^{-9} g}
1  mg=106  ng\Rightarrow 1 \;mg = 10^ 6\; ng
1  mg=106  kg=106  ng1 \;mg = 10 ^{-6}\; kg = 10 ^6 \; ng

(iii) 1  mL=1  mL×1L1000mL1\; mL = 1 \;mL × \frac{1 \,L }{ 1000 \,mL}
1  mL=103  L\Rightarrow 1 \;mL = 10 ^{-3}\; L
1  mL=1  cm3=1  cm31dm×1dm×1dm10cm×10cm×10cm1\; mL = 1\; cm^3 = 1 \;cm^3 \frac{1 dm × 1 dm × 1 dm}{ 10 cm × 10 cm × 10 cm}
1  mL=103dm3\Rightarrow 1\; mL = 10 ^{-3} dm^3
\therefore 1  mL=103  L=103  dm31\; mL = 10 ^{-3}\; L = 10 ^{-3} \;dm^3