SELECT * FROM question_mgmt as q WHERE id=1237 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=42 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 11th chemistry Some Basic Concepts of Chemistry the following data are obtained when dinitrogen an

Question:

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

 

(a) Which law of chemical combination is obeyed by the above experimental data?Give its statement.

 

(b) Fill in the blanks in the following conversions:

(i) 1 km = ...................... mm = ...................... pm

(ii) 1 mg = ...................... kg = ...................... ng

(iii) 1 mL = ...................... L = ...................... dm3

Answer:

Let us fix 14 parts by weight of nitrogen as fixed weight.

 

Now let us calculate the weights of oxygen which combine with 14 parts by weight of nitrogen

 

Sno

No of parts by weight of nitrogen

No of parts by weight of oxygen

14 parts of nitrogen as fixed weight

No of parts by weight of oxygen which combine with 14 parts by weight of nitrogen

1

14g

16g

14g

16

2

14g

32g

14g

32

3

28g

32g

14g

32

4

28g

80g

14g

80

 

(a)  If we fix the mass of dinitrogen at 14 g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 16 g, 32 g, 32 g, and 80 g.

     The masses of dioxygen bear a whole number ratio of 1:2:2:5. Hence, the given experimental data obeys the law of multiple proportions.

     This law was given by Dalton in 1804. The law states that if two elements combine to form 2 or more compound, then the weight of one element which            combines a  fixed weight  of other element in these compounds,bears a simple whole number ratio by weight.

(b) (i) We know 1km=1000m

      Or 1m = 1000 mm

      Therefore 1km = 1000x 1000mm= 106 mm

      1 km = 1 km ×

      1 km = 1015 pm 

      Hence, 1 km = 106 mm = 1015 pm

     (ii)  We know 1kg = 1000mg

          Or 1000mg= 1kg

          Or 1mg= 1/1000* 1= 0.01 kg

          1 mg = 1 mg × 

          ⇒ 1 mg = 106 ng

             1 mg = 10–6 kg = 106 ng

     (iii)  We know 1000 ml=l L

            Or 1ml=1/1000*1= 0.01L

            1 mL = 1 cm3 = 1 cm3

            ⇒ 1 mL = 10–3 dm3

            1 mL = 10–3 L = 10–3 dm3

   

 


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