Answer:
30 g 0f NaOH.
Explanation:
The following data were obtained from the question:
Molarity of NaOH = 0.25 M
Volume = 3 dm³
Mass of NaOH =?
Next, we shall determine the number of mole of NaOH present in 3 dm³ of 0.25 M NaOH solution.
This can be obtained as follow:
Molarity of NaOH = 0.25 M
Volume = 3 dm³
Mole of NaOH =?
Molarity = mole /Volume
0.25 = mole of NaOH/3
Cross multiply
Mole of NaOH = 0.25 × 3
Mole of NaOH = 0.75 mole
Finally, we shall determine the mass of NaOH as follow:
Mole of NaOH = 0.75 mole
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
Mass of NaOH =.?
Mole = mass /Molar mass
0.75 = mass of NaOH /40
Cross multiply
Mass of NaOH = 0.75 × 40
Mass of NaOH = 30 g
Therefore, 30 g 0f NaOH is present in the flask.
The mass of NaOH in 0.25 M sample has been [tex]\rm\bold{ 30\;g/dm^3}[/tex].
The molarity has been defined as moles of sample present in a liter of solution.
The molarity of given NaOH solution has been 0.25 M. It has been stating that 0.25 moles of NaOH has been present in a liter of solution.
The mass of sample present in 1 cubic decimeter has been the mass of sample in a liter of solution as:
[tex]\rm 1\;dm^3=1\;L[/tex]
The molarity (M) has been given as:
[tex]M=\dfrac{\text {mass}}{mwt}\;\times\;\dfrac{1}{V(L)}[/tex]
Where, the molarity of NaOH sample has been, [tex]M=0.25\;\text M[/tex]
The molar mass of NaOH has been, [tex]mwt=40\;\text {g/mol}[/tex]
The volume of solution has been, [tex]V=3 \;\text L[/tex]
Substituting the values for mass of NaOH:
[tex]0.25=\dfrac{mass}{40}\;\times\;\dfrac{1}{3} \\mass=0.25\;\times\;40\;\times\;3\\mass=30\;\text g[/tex]
The mass of NaOH in 0.25 M sample has been [tex]\rm\bold{ 30\;g/dm^3}[/tex].
For more information about molarity, refer to the link:
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