Question

Match accordingly

16) The molarity of 0.15 L containing 0.5 moles of NaCl                                  a. 1.2 L

17) The molarity of 0.289 moles of FeCl3 dissolved in 120 mls solution     b. 3.4 M

18) The molarity in 650 ml solution containing 63 grams NaCl                     c. 2.41 M

19) The amount of a 0.88 M solution made using 130 grams of FeCl2       d. 1.7 M

Match accordingly to plant tissue culture

20) Placing sterile plant back into soil                      a. stage 1

21) High levels of auxin                                                  b. stage 2

22) Placing sterile explant on media                         c. stage 0

23) High levels of cytokinin to promote shoots    d. stage 4

24) Mother plant                                                                              e. stage 3

Answer 1

16)--b is correct because molarity is the moles of solute divided by litres of solution .So 0.5/0.15=3.3M.

17)-option c is correct because firstly we convert 120mls into 120 litre.120mls=0.12L.So 0.289/0.12=2.41M.

18)-option d is correct because firstly we convert 63 gram of NaCl into moles.63g =1.078moles.So 1.078moles/0.65L=1.7M.

20)-option d is correct because it is last stage (hardening).

21)- option e is correct because auxin initiate root formation.

22)--option a is correct because explant inoculate on media for proper growth.

23)-option b is correct because cytokinin initiate shoot formation.

24)-option c is correct because this is the first stage in which we check mother plant.