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Question

(B) The molar mass of $NaOH = 23 + 16 + 1 = 40 \, g \, mol^{-1}$.Moles of $NaOH = \frac{	ext{mass}}{	ext{molar mass}} = \frac{5 \, g}{40 \, g \, mol

Vedclass · Accepted Answer

$0.278$

Vedclass · Answer

(B) The molar mass of $NaOH = 23 + 16 + 1 = 40 \, g \, mol^{-1}$.Moles of $NaOH = \frac{	ext{mass}}{	ext{molar mass}} = \frac{5 \, g}{40 \, g \, mol^{-1}} = 0.125 \, mol$.Volume of the solution in litres $= \frac{450 \, mL}{1000 \, mL \, L^{-1}} = 0.450 \, L$.Molarity $(M) = \frac{	ext{moles of solute}}{	ext{volume of solution in litres}} = \frac{0.125 \, mol}{0.450 \, L} = 0.2777... \, M \approx 0.278 \, M$.

Calculate the molarity of a solution containing $5 \, g$ of $NaOH$ in $450 \, mL$ solution. (in $, M$)

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