2-0 g of polybasic organic acid molecular wt- - 600 require 100 mL of a M6 NaOH solution for complete neutralisation- The basicity of the acid is-

No. of equivalents of acid = No. of equivalents of base mass × basicity(n)Molar mass = volume (L) × Normality 2× basicity(n)600 = 0.100×16 ∴ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Stoichiometric Calculations

2.0 g of poly...

Question

$2.0 g$ of polybasic organic acid (molecular wt. $= 600$ ) require $100 m L$ of a $\frac{M}{6} N a O H$ solution for complete neutralisation. The basicity of the acid is:

Open in App

Solution

No. of equivalents of acid $=$ No. of equivalents of base
$\frac{m a s s \times b a s i c i t y (n)}{M o l a r m a s s} = v o l u m e (L) \times N o r m a l i t y$

$\frac{2 \times b a s i c i t y (n)}{600} = 0.100 \times \frac{1}{6}$
$∴ B a s i c i t y (n) = 5$ .

Suggest Corrections

Similar questions

1.575

gm of an organic acid was dissolved in

250

ml of water. Further,

20

ml of this solution required

16

ml of

\frac{N}{8}

alkali solution for complete neutralisation. If the basicity of the acid is

2

, find its molecular mass.

0.16 g

a dibasic acid required

25 m L

of decinormal

N a O H

solution for complete neutralisation. The molecular mass of the acid is:

0.16g of dibasic acid required 25 ml of decinprmal NaOH solution for complete neutralisation. The molecular weight of acid will be

A known mass of oxalic acid dihydrate is dissolved in water to make 100 ml solution. 10 ml of this solution is diluted to 100 ml. 20 mL of the solution requires 20 mL of 0.1 N NaOH for complete neutralisation. (eq. mass of oxalic acid dihydrate = 63). The mass of oxalic acid in gram in the original solution is

0.4

g of polybasic acid,

H_{n} A

(all the hydrogens are acidic) requires

0.5

g of

N a O H

for complete neutralization.
The number of replaceable hydrogen atoms and the molecular weight of

A

would be respectively : (molecular weight of acid

= 96

g/mol)

Join BYJU'S Learning Program

2.0g of polybasic organic acid (molecular wt. =600) require 100mL of a M6 NaOH solution for complete neutralisation. The basicity of the acid is:

No. of equivalents of acid = No. of equivalents of basemass×basicity(n)Molar mass=volume(L)×Normality2×basicity(n)600=0.100×16∴Basicity(n)=5.

$2.0 g$ of polybasic organic acid (molecular wt. $= 600$ ) require $100 m L$ of a $\frac{M}{6} N a O H$ solution for complete neutralisation. The basicity of the acid is:

No. of equivalents of acid $=$ No. of equivalents of base
$\frac{m a s s \times b a s i c i t y (n)}{M o l a r m a s s} = v o l u m e (L) \times N o r m a l i t y$

$\frac{2 \times b a s i c i t y (n)}{600} = 0.100 \times \frac{1}{6}$
$∴ B a s i c i t y (n) = 5$ .