The normality of a solution that results from mixing 4 g of NaOH , 500 mL of 1 M HCl, and 10.0 mL 2 of H 2 O 4 (specific gravity 1.1,49 % H 2 SO 4 by weight) is

Question

The normality of a solution that results from mixing 4 g of NaOH , 500 mL of 1 M HCl, and 10.0 mL 2 of H 2 O 4 (specific gravity 1.1,49 % H 2 SO 4 by weight) is (A) 0.51 (B) 0.71 (C) 1.02 (D) 0.45

Tardigrade Admin · Accepted Answer

4 g of NaOH =4 / 40=0.1 mole =100 mmol ≡ 100 mEq HCl =500 × 1=500 mEq N of H 2 SO 4=(W2 × 1000/E w2 × V text sol ( in mL )) or =( % text by weight × 10 × d/E w2) =(49 × 10 × 1.1/49)=11 N mEq of H 2 SO 4=11 N × 10.00 mL =110 mEq Total acid =100+500=610 mEq NaOH =100 mEq Acid left =610-100=510 mEq Total volume =1000 mL Normality of solution =( mEq / mL )=(510/1000)=0.51 N

Q. The normality of a solution that results from mixing $4 g$ of $N a O H, 500 m L$ of $1 M H Cl$ , and $10.0 m L^{2}$ of $H_{2} O_{4}$ (specific gravity $1.1, 49% H_{2} S O_{4}$ by weight) is

$0.51$

$0.71$

$1.02$

$0.45$

Q. The normality of a solution that results from mixing 4g of NaOH,500mL of 1MHCl, and 10.0mL2 of H2​O4​ (specific gravity 1.1,49%H2​SO4​ by weight) is

0.51

0.71

1.02

0.45

Q. The normality of a solution that results from mixing $4 g$ of $N a O H, 500 m L$ of $1 M H Cl$ , and $10.0 m L^{2}$ of $H_{2} O_{4}$ (specific gravity $1.1, 49% H_{2} S O_{4}$ by weight) is

$0.51$

$0.71$

$1.02$

$0.45$