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,

Question

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)

Prashant Kotian · Accepted Answer

Correct answer: .. 2Na + 2H2O  2NaOH + H2 Number of moles of Na = No. of moles of NaOH nNaOH = 0.69 23 = 0.03 NaOH + HCl  NaCl + H2O Moles of HCl = moles of NaOH (Molarity × V)HCl = Number of moles of NaOH       1 . 73 × V = 0.03 V = 15 ×10–3Lit = 15 mL.

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