0.45 g of acid molecular weight 90 is neutralised by 20 ml of 0.5 N caustic potash. The basicity of acid is

Question

0.45 g of acid molecular weight 90 is neutralised by 20 ml of 0.5 N caustic potash. The basicity of acid is (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

Let the n text factor be n. Moles of acid =(0.45/90)=5 milimoles Comparing equivalents of acid and NaOH 5 × n=0.5 × 20 n =2 For acid, n text factor = basicity ∴ Basicity is 2

Q. $0.45 g$ of acid molecular weight $90$ is neutralised by $20 m l$ of $0.5 N$ caustic potash. The basicity of acid is

1

2

3

4

Let the $n_{factor}$ be $n$ .
Moles of acid $= \frac{0.45}{90} = 5$ milimoles
Comparing equivalents of acid and $N a O H$
$5 \times n = 0.5 \times 20$
$n = 2$
For acid, $n_{factor} =$ basicity
$∴$ Basicity is $2$

Q. 0.45g of acid molecular weight 90 is neutralised by 20ml of 0.5N caustic potash. The basicity of acid is

1

2

3

4

Let the nfactor ​ be n. Moles of acid =900.45​=5 milimoles Comparing equivalents of acid and NaOH 5×n=0.5×20 n=2 For acid, nfactor ​= basicity ∴ Basicity is 2

Q. $0.45 g$ of acid molecular weight $90$ is neutralised by $20 m l$ of $0.5 N$ caustic potash. The basicity of acid is

Let the $n_{factor}$ be $n$ .
Moles of acid $= \frac{0.45}{90} = 5$ milimoles
Comparing equivalents of acid and $N a O H$
$5 \times n = 0.5 \times 20$
$n = 2$
For acid, $n_{factor} =$ basicity
$∴$ Basicity is $2$