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

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The reaction is

\mathrm{Na_{2}O+H_{2}O\rightarrow 2NaOH}

Moles of  \mathrm{Na_{2}O=\frac{20}{62}=\frac{10}{31}}

\therefore Moles of  \mathrm{NaOH} obtained  = \frac{20}{31}  in  \mathrm{500\ ml}

\mathrm{\therefore \left [ NaOH \right ]=\frac{20}{31}\times2=\frac{40}{31}=1.29=12.9 \times 10^{-1}}

Hence, the answer is 13

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sudhir.kumar

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