(Assume one mole of gas occupies 24dm^3 at rt.p.) [1] (b) (i) A student completely decomposed 3.25 g of sodium azide. Calculate the mass of sodium she obtained __ 2 (ii) She then carefully reacted the sodium obtained with water to form 25.0cm^3 of aqueous sodium hydroxide 2Na(s)+2H_(2)O(l)arrow 2NaOH(aq)+H_(2)(g) Calculate the concentration, in moldm^-3 of the aqueous sodium hydroxide. [2] (Total for Question 11=5 marks

(b) (i) To calculate the mass of sodium obtained from the decomposition of sodium azide, we need to use the balanced chemical equation for the decomposition reaction. The balanced equation is:<br /><br />$2NaN_{3}(s) \rightarrow 2Na(s) + 3N_{2}(g)$<br /><br />From the balanced equation, we can see that 2 moles of sodium azide decompose to produce 2 moles of sodium. Therefore, the moles of sodium obtained from the decomposition of 3.25 g of sodium azide can be calculated using the molar mass of sodium azide (65.01 g/mol) and the molar mass of sodium (22.99 g/mol).<br /><br />Moles of sodium azide = $\frac{3.25 \text{ g}}{65.01 \text{ g/mol}}$<br /><br />Moles of sodium = $\frac{3.25 \text{ g}}{65.01 \text{ g/mol}} \times \frac{2 \text{ moles of sodium azide}}{2 \text{ moles of sodium}}$<br /><br />Mass of sodium = Moles of sodium $\times$ Molar mass of sodium<br /><br />(ii) To calculate the concentration of the aqueous sodium hydroxide, we need to use the balanced chemical equation for the reaction between sodium and water. The balanced equation is:<br /><br />$2Na(s) + 2H_{2}O(l) \rightarrow 2NaOH(aq) + H_{2}(g)$<br /><br />From the balanced equation, we can see that 2 moles of sodium react with 2 moles of water to produce 2 moles of sodium hydroxide. Therefore, the moles of sodium hydroxide produced from the reaction of the sodium obtained can be calculated using the moles of sodium obtained from part (i) and the stoichiometry of the reaction.<br /><br />Moles of sodium hydroxide = Moles of sodium obtained from part (i)<br /><br />Concentration of sodium hydroxide = $\frac{\text{Moles of sodium hydroxide}}{\text{Volume of aqueous sodium hydroxide in dm}^3}$<br /><br />Volume of aqueous sodium hydroxide = $25.0 \text{ cm}^3 = 0.025 \text{ dm}^3$<br /><br />Concentration of sodium hydroxide = $\frac{\text{Moles of sodium hydroxide}}{0.025 \text{ dm}^3}$<br /><br />Please note that the specific calculations for the mass of sodium and the concentration of sodium hydroxide would require numerical values and units to be provided in the question.