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

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)

TestHub · Accepted Answer

Mole of NaOH = 0 . 69 23 = 3 × 10 - 2 2 Na + 2 H 2 O ⟶ 2 NaOH + H 2 Number of moles of Na = Number of moles of NaOH NaOH + HCl → NaCl + H 2 O Moles of HCl = moles of NaOH ⇒ ( Molarity × V ) HCl = Number of moles of NaOH ⇒ 73 36 . 5 × 1 × V = 0 . 03 ⇒ V = 15 × 10 - 3 Lit ⇒ V = 15 mL

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