Answer: -
3.151 M
Explanation: -
Let the volume of the solution be 1000 mL.
At 25.0 °C, Density = 1.260 g/ mL
Mass of the solution = Density x volume
= 1.260 g / mL x 1000 mL
= 1260 g
At 25.0 °C, the molarity = 3.179 M
Number of moles present per 1000 mL = 3.179 mol
Strength of the solution in g / mol
= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)
Now at 50.0 °C
The density is 1.249 g/ mL
Mass of the solution = density x volume = 1.249 g / mL x 1000 mL
= 1249 g.
Number of moles present in 1249 g = Mass of the solution / Strength in g /mol
= 
= 3.151 moles.
So 3.151 moles is present in 1000 mL at 50.0 °C
Molarity at 50.0 °C = 3.151 M
A bag of soil will weigh in kilograms
Explanation:
As the total concentration is given as 1.2 mM. And, it is also given that salt present in the solution is NaCl.
As sodium chloride is an ionic compound so, when it is added to water then it will dissociate into sodium and chlorine ions as follows.

So, it means in total there will be formation of 2 ions when one molecules of NaCl dissociates.
Therefore, concentration of chlorine ions will be calculated as follows.
Concentration of
ions =
= 0.6 mM
Thus, we can conclude that the concentration of chloride ions is 0.6 mM.
2NaOH=Na2O +H2O
The coefficient of NaOH is 2