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

Name: How many grams of dibasic acid (mol. wt. 200) should be present in 100 mL of the aqueous solution to give 0.1 N ?
Uploaded: 2020-01-06
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Question

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Answer 1

Equivalent weight of dibasic acid
$= \frac{m o l , w t .}{2} = \frac{2 0 0}{2} = 100$

Strength = 0.1 N, m = ?, V= 100 ml.

From, $m = \frac{E V N}{1 0 0 0} = \frac{1 0 0 \times 1 0 0 \times 0 . 1}{1 0 0 0} = 1 g$

Hence, the correct option is $A$

Answer 2

Equivalent weight of dibasic acid

= \frac{m o l , w t .}{2} = \frac{2 0 0}{2} = 100

Strength = 0.1 N, m = ?, V= 100 ml.

From,

m = \frac{E V N}{1 0 0 0} = \frac{1 0 0 \times 1 0 0 \times 0 . 1}{1 0 0 0} = 1 g

Hence, the correct option is

A

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

1 g

2 g

10 g

20 g

Equivalent weight of dibasic acid
$= \frac{m o l , w t .}{2} = \frac{2 0 0}{2} = 100$

Strength = 0.1 N, m = ?, V= 100 ml.

From, $m = \frac{E V N}{1 0 0 0} = \frac{1 0 0 \times 1 0 0 \times 0 . 1}{1 0 0 0} = 1 g$

Hence, the correct option is $A$

250 ml of a solution contains 6.3 grams of oxalic acid(mol. wt. =126). What is the volume (in litres) of water to be added to this solution to make it a 0.1 N solution?

Weight of solute present in 500 ml of 0.2 $N H_{2} S O_{4}$ solution is:

If $4$ g NaOH are dissolved in $100$ mL of aqueous solution, what will be the difference in its normality and molarity?

$0.1 N$ solution of a dibasic acid can be prepared by dissolving $0.45 g$ of the acid in water and diluting to $100 m L$ . The molecular mass of the acid is:

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

1 g

2 g

10 g

20 g

Equivalent weight of dibasic acid =2mol,wt.​=2200​=100 Strength = 0.1 N, m = ?, V= 100 ml. From, m=1000E V N​=1000100×100×0.1​=1g Hence, the correct option is A

250 ml of a solution contains 6.3 grams of oxalic acid(mol. wt. =126). What is the volume (in litres) of water to be added to this solution to make it a 0.1 N solution?

Weight of solute present in 500 ml of 0.2N H2​SO4​ solution is:

If 4 g NaOH are dissolved in 100 mL of aqueous solution, what will be the difference in its normality and molarity?

0.1 N solution of a dibasic acid can be prepared by dissolving 0.45 g of the acid in water and diluting to 100 mL. The molecular mass of the acid is:

Equivalent weight of dibasic acid
$= \frac{m o l , w t .}{2} = \frac{2 0 0}{2} = 100$

Strength = 0.1 N, m = ?, V= 100 ml.

From, $m = \frac{E V N}{1 0 0 0} = \frac{1 0 0 \times 1 0 0 \times 0 . 1}{1 0 0 0} = 1 g$

Hence, the correct option is $A$

Weight of solute present in 500 ml of 0.2 $N H_{2} S O_{4}$ solution is:

If $4$ g NaOH are dissolved in $100$ mL of aqueous solution, what will be the difference in its normality and molarity?

$0.1 N$ solution of a dibasic acid can be prepared by dissolving $0.45 g$ of the acid in water and diluting to $100 m L$ . The molecular mass of the acid is: