How many g of dibasic acid (mol. wt. 200) should be present in 100 mL of the aqueous solution to give 0.1 normality ?

Question

How many g of dibasic acid (mol. wt. 200) should be present in 100 mL of the aqueous solution to give 0.1 normality ? (A) 1 g (B) 2 g (C) 10 g (D) 20 g

Tardigrade Admin · Accepted Answer

Equivalent weight of dibasic acid =( mol , text wt. /2)=(200/2)=100 Strength =0.1 N , m =?, V =100 ml . From, m =( E V N /1000) =(100 × 100 × 0.1/1000)=1 g

Q. How many g of dibasic acid (mol. wt. 200) should be present in 100 mL of the aqueous solution to give 0.1 normality ?

1 g

2 g

10 g

20 g

Equivalent weight of dibasic acid
$= \frac{m o l , wt.}{2} = \frac{200}{2} = 100$
Strength $= 0.1 N, m = ?, V = 100 m l .$
From, $m = \frac{E V N}{1000}$
$= \frac{100 \times 100 \times 0.1}{1000} = 1 g$

Q. How many g of dibasic acid (mol. wt. 200) should be present in 100 mL of the aqueous solution to give 0.1 normality ?

1 g

2 g

10 g

20 g

Equivalent weight of dibasic acid =2mol, wt. ​=2200​=100 Strength =0.1N,m=?,V=100ml. From, m=1000EVN​ =1000100×100×0.1​=1g

Equivalent weight of dibasic acid
$= \frac{m o l , wt.}{2} = \frac{200}{2} = 100$
Strength $= 0.1 N, m = ?, V = 100 m l .$
From, $m = \frac{E V N}{1000}$
$= \frac{100 \times 100 \times 0.1}{1000} = 1 g$