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PART - 1 CH - 1 Some Basic Concepts of Chemistry — Chemistry STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistryPART - 1 CH - 1 Some Basic Concepts of Chemistry3 Marks
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(a) Taking a fixed mass of dinitrogen, say 28 gm , the mass of dioxygen that reacts with it to form oxide is $32,64,32$ and 80 respectively, whose simple ratio is $2: 4: 2: 5$ which is a simple integer ratio. Therefore, these data are in accordance with law of multiple proportion.
(b) (i) $\begin{aligned} 1 km=1 km \times \frac{1000 m}{1 km} \times \frac{100 cm}{1 m} & \times \frac{10 mm}{1 cm} \\ & =10^6 mm\end{aligned}$
$\begin{array}{l}1 km=1 km \times \frac{1000 m}{1 km} \times \frac{1 pm }{10^{-12} m}=10^{15} Pm \\ \text { Hence } 1 km=10^6 mm=10^{15} pm \end{array}$
(ii) $1 mg =\frac{1 mg \times 1 g}{1000 mg } \times \frac{1 kg}{1000 g}=10^{-6} kg$
$\begin{array}{l}1 mg =1 mg \times \frac{1 g}{1000 mg } \times \frac{1 ng }{10^{-9} g}=10^6 ng \\ \text { Hence } 1 mg =10^{-6} kg=10^6 ng \end{array}$
$\begin{array}{l}\text { (iii) } 1 mL=1 mL \times \frac{1 L}{1000 mL}=10^{-3} L \\ 1 mL=1 cm^3=1 cm^3 \times \frac{1 dm }{10 cm} \times \frac{1 dm }{10 cm} \\ \times \frac{1 dm }{10 cm}=10^{-3} dm ^3 \\ \text { Hence } 1 mL=10^{-3} L=10^{-3} dm ^3\end{array}$

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