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
Answer:
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†•°-Love Ash or Ashlynn-°•†
P.S (Have A great day!!)
Molar mass of LiBr (mm )= 86.845 g/mol
Molarity ( M ) = 4 M
Mass of solute ( m ) = 100 g
Volume ( V ) = in liters ?
V = m / mm * M
V = 100 / 86.845 * 4
V = 100 / 347.38
V = 0.2875 L
hope this helps!.
Answer:
The third one makes the most sense