Match accordingly 16) The molarity of 0.15 L containing 0.5 moles of NaCl a. 1.2
ID: 275146 • Letter: M
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
Explanation / Answer
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.