The volume of HCl, containing $73 g L^{−1}$, required to completely neutralise NaOH obtained by reacting $0.69 g$ of metallic sodium with water, is _______ mL.( Nearest Integer) (Given : molar Masses of Na, Cl, O, H, are 23,35.5,16 and 1 $g mol^{−1}$ respectively)
First, we determine the amount of NaOH formed. The reaction of sodium with water is as follows: 2 Na + 2 H2O → 2 NaOH + H2 One mole of Na (23 g) produces one mole of NaOH (40 g), as the molar masses are Na = 23 g/mol and NaOH = 40 g/mol. The mass of sodium given is 0.69 g. Calculate moles of Na: moles Na = 0.69 g ÷ 23 g/mol = 0.03 mol Therefore, moles of NaOH formed = 0.03 mol (1:1 ratio with Na). Next, calculate the volume of HCl needed for neutralization: HCl reacts with NaOH in a 1:1 molar ratio: NaOH + HCl → NaCl + H2O The concentration of HCl is 73 g/L. Find its molarity: Molar mass of HCl = 36.5 g/mol Molarity HCl = 73 g/L ÷ 36.5 g/mol = 2 mol/L The volume of HCl required to neutralize 0.03 mol of NaOH: Volume HCl = moles of NaOH ÷ molarity of HCl = 0.03 mol ÷ 2 mol/L = 0.015 L Convert volume in liters to milliliters: 0.015 L × 1000 mL/L = 15 mL The final volume of HCl required to neutralize the NaOH is 15 mL, which fits within the given range.
