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(B) The balanced chemical equation is: $Na_{2}O + H_{2}O ightarrow 2 NaOH$Molar mass of $Na_{2}O = (2 	imes 23.0) + 16.0 = 62.0 \ g/mol$.Moles of $

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$13$

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(B) The balanced chemical equation is: $Na_{2}O + H_{2}O ightarrow 2 NaOH$Molar mass of $Na_{2}O = (2 	imes 23.0) + 16.0 = 62.0 \ g/mol$.Moles of $Na_{2}O = \frac{20.0 \ g}{62.0 \ g/mol} = 0.3226 \ mol$.According to the stoichiometry,$1 \ mol$ of $Na_{2}O$ produces $2 \ mol$ of $NaOH$.Moles of $NaOH$ formed $= 2 	imes 0.3226 \ mol = 0.6452 \ mol$.Volume of solution $= 500 \ mL = 0.5 \ L$.Molarity of $NaOH = \frac{0.6452 \ mol}{0.5 \ L} = 1.2904 \ M$.Expressing in terms of $10^{-1} \ M$: $1.2904 \ M = 12.904 	imes 10^{-1} \ M$.Rounding to the nearest integer,we get $13 	imes 10^{-1} \ M$.

Sodium oxide reacts with water to produce sodium hydroxide. $20.0 \ g$ of sodium oxide is dissolved in $500 \ mL$ of water. Neglecting the change in volume,the concentration of the resulting $NaOH$ solution is $........ \times 10^{-1} \ M.$ (Nearest integer)
[Atomic mass $: Na=23.0, O=16.0, H=1.0]$

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Sodium oxide reacts with water to produce sodium hydroxide. $20.0 \ g$ of sodium oxide is dissolved in $500 \ mL$ of water. Neglecting the change in volume,the concentration of the resulting $NaOH$ solution is $........ \times 10^{-1} \ M.$ (Nearest integer)[Atomic mass $: Na=23.0, O=16.0, H=1.0]$