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Some Basic Concepts of Chemistry — Chemistry STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistrySome Basic Concepts of Chemistry3 Marks
Question
The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
Fill in the blanks in the following conversions:
  1. 1km = ...................... mm = ...................... pm.
  2. 1mg = ...................... kg = ...................... ng.
  3. 1mL = ...................... L = ...................... dm3.
✓

Answer

  1. $1\text{km}= 1\text{km}\times\frac{1000\text{m}}{1\text{km}}\times\frac{100\text{cm}}{1\text{m}}\times\frac{10\text{mm}}{1\text{cm}}$

$\therefore1\text{km}=10^6\text{mm}$

$1\text{km}= 1\text{km}\times\frac{1000\text{m}}{1\text{km}}\times\frac{1\text{pm}}{10^{-12}\text{m}}$

$\therefore1\text{km}=10^{15}\text{pm}$

Hence, $1\text{km} = 10^6\text{mm} = 10^{15}\text{pm}$

  1. $1\text{mg} = 1\text{mg}×\frac{1\text{g}}{1000\text{mg}}\times\frac{1\text{kg}}{1000\text{g}}$

$⇒ 1 \text{mg} = 10^{–6} \text{kg}$

$1\text{mg} = 1\text{mg}×\frac{1\text{g}}{1000\text{mg}}\times\frac{1\text{ng}}{10^{-9}\text{g}}$

$⇒ 1 \text{mg} = 10^{–6} \text{ng}$

$\therefore1\text{mg}=10^{-6}\text{kg}=10^{6}\text{ng}$

  1. $1 \text{mL} = 1 \text{mL} × \frac{1\text{L}}{1000\text{mL}}$ 

$⇒ 1 \text{mL} = 10^{–3}\text{L}$

$1 \text{mL} = 1 \text{cm}^3 = 1 \text{cm}^3\frac{1\text{dm}\times1\text{dm}\times1\text{dm}}{10\text{cm}\times10\text{cm}\times10\text{cm}}$

$⇒ 1 \text{mL} = 10^{–3}\text{dm}^3$

$\therefore 1 \text{mL} = 10^{–3} \text{L} = 10^{–3} \text{dm}^3$

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