what will be the percentage composition by mass of a solution made by dissolving 22.5 g of sodium hydroxide in 75.0 g water

We are Given:

22.5 grams of NaOH in 75 grams of H₂O

2NaOH + 2H₂O →  2Na(OH)₂ + H₂

from this equation, we can see that for 2 moles of both NaOH and H₂O, we get 2 Moles of Na(OH)₂

Calculating the number of moles of NaOH and H₂O provided:

Moles of NaOH:

molar mass of NaOH = 40 grams/mol

Given mass of NaOH = 22.5 grams

Number of moles of NaOH = Given mass / Molar mass

Number of moles of NaOH = 22.5 / 40

Number of moles of NaOH = 1.125 moles of NaOH

Moles of H₂O:

molar mass of H₂O = 18 grams / mol

given mass of H₂O = 75 grams

Number of moles of H₂O = Given mass / Molar mass

Number of moles of H₂O = 75 / 18 = 4.167 moles of H₂O

Number of Moles of NaOH and H₂O which will actually react:

from the equation, we know that the moles of NaOH and H₂O react in 1:1 ratio

So, we can say that one mole of NaOH reacts with one mole of H₂O

but we have much less NaOH than H₂O

So, only the moles equal to the number of moles of NaOH will react of either compound

1.125 moles of both NaOH and H₂O will react

Moles of Na(OH)₂ formed:

we know from the equation that the moles of NaOH and H₂O react to form  moles of Na(OH)₂ in the ratio 1:1:1

So, 1.125 moles of Na(OH)₂ will be formed

Percentage composition by mass:

amount of water present alongside the Na(OH)₂ formed:

we were given 4.167 moles of H₂O and only 1.125 moles of it was consumed in the reaction

so, 4.167 - 1.125 = 3 moles of H₂O will be present alongside the Na(OH)₂

3 moles of H₂O = 3 * 18 = 54 grams of H₂O will be present along the Na(OH)₂

Mass of Na(OH)₂ formed:

mass = number of moles * molar mass

mass = 1.125 * 41  = 46 grams of Na(OH)₂ will be formed

Total mass of the solution formed:

total mass = mass of H₂O + mass of Na(OH)₂

total mass = 54 + 46 = 100 grams

Percent composition by mass of Na(OH)₂:

Mass% of Na(OH)₂ = (Mass of Na(OH)₂ / Total mass) * 100

Mass% of Na(OH)₂ = 46 %

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