A flask contains 0.25 naoh. what mass of naoh is present per dm3 of solution

The mass of NaOH in 0.25 M sample has been .

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:

The molarity (M) has been given as:

Where, the molarity of NaOH sample has been,

The molar mass of NaOH has been,

The volume of solution has been,

Substituting the values for mass of NaOH:

The mass of NaOH in 0.25 M sample has been .

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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.