0.45 g of an acid of molecular weight 90 was neutralized by 20 mL of a 0.5 N caustic potash. The basicity of an acid is

Question

0.45 g of an acid of molecular weight 90 was neutralized by 20 mL of a 0.5 N caustic potash. The basicity of an acid is (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

Equivalents of KOH used =(20 × 0.5/1000)=0.01 Moles of acid =(0.45/90)=0.005 mol If basicity of acid = x Then equivalents of acid =0.005 x ∴ 0.005 x =0.01 ∴ x=2

Q. $0.45 g$ of an acid of molecular weight $90$ was neutralized by $20 m L$ of a $0.5 N$ caustic potash. The basicity of an acid is

1

2

3

4

Equivalents of $K O H$ used $= \frac{20 \times 0.5}{1000} = 0.01$
Moles of acid $= \frac{0.45}{90} = 0.005 m o l$
If basicity of acid $= x$
Then equivalents of acid $= 0.005 x$
$∴ 0.005 x = 0.01$
$∴ x = 2$

Q. 0.45g of an acid of molecular weight 90 was neutralized by 20mL of a 0.5N caustic potash. The basicity of an acid is

1

2

3

4

Equivalents of KOH used =100020×0.5​=0.01 Moles of acid =900.45​=0.005mol If basicity of acid =x Then equivalents of acid =0.005x ∴0.005x=0.01 ∴x=2

Q. $0.45 g$ of an acid of molecular weight $90$ was neutralized by $20 m L$ of a $0.5 N$ caustic potash. The basicity of an acid is

Equivalents of $K O H$ used $= \frac{20 \times 0.5}{1000} = 0.01$
Moles of acid $= \frac{0.45}{90} = 0.005 m o l$
If basicity of acid $= x$
Then equivalents of acid $= 0.005 x$
$∴ 0.005 x = 0.01$
$∴ x = 2$