The volume of HCl, containing 73g L^(-1), required to completely neutralise NaOH obtained by reacting 0.69 g of metallic sodium with water, is _____…

Question

The volume of HCl, containing 73g 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 1g mol^(-1) respectively)

MARKS Content Team · Accepted Answer

Correct answer: 15. Mole of NaOH=0.69/23=3×10^(-2) 2Na+2H_2O⟶2NaOH+H_2 Number of moles of Na = Number of moles of NaOH NaOH + HCl → NaCl + H_2O 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

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)

Solution

Relevant Questions

The volume of HCl, containing 73g 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 1g mol-1 respectively)

Solution

Relevant Questions

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)