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)

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) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Mole of Na =(0.69/23)=3 × 10-2 Na + H 2 O longrightarrow NaOH +(1/2) H 2 By using POAC Moles of NaOH =3 × 10-2 NaOH reacts with HCl No. of equivalent of NaOH = No. of equivalent of HCl 3 × 10-2 × 1=(73/36.5) × V ( text in L ) × 1 V =1.5 × 10-2 L Volume of HCl =15 ml.

Q. The volume of HCl, containing 73gL−1, required to completely neutralise NaOH obtained by reacting 0.69g of metallic sodium with water, is-__ mL. (Nearest Integer) (Given : molar Masses of Na,Cl,O,H are 23,35.5,16 and 1gmol−1 respectively)

Mole of Na=230.69​=3×10−2 Na+H2​O⟶NaOH+21​H2​ By using POAC Moles of NaOH=3×10−2 NaOH reacts with HCl No. of equivalent of NaOH=No. of equivalent of HCl 3×10−2×1=36.573​×V( in L)×1 V=1.5×10−2L Volume of HCl=15ml.

Q. The volume of $H Cl$ , containing $73 g L^{- 1}$ , required to completely neutralise $N a O H$ obtained by reacting $0.69 g$ of metallic sodium with water, is-__ $m L$ . (Nearest Integer)
(Given : molar Masses of $N a, Cl, O, H$ are $23, 35.5, 16$ and $1 g m o l^{- 1}$ respectively)