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

Question

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

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

Video Explanation

Answer

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 $A$

Practice important Questions

BCECE 2012 Medical Chemistry

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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

Video Explanation

Answer

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

Practice important Questions

BCECE 2012 Medical Chemistry

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Take Toppr Scholastic Test for Aptitude and Reasoning

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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 $A$

The normality of $0.3$ M phosphorus acid $(H_{3} P O_{3})$ solution is:

Normality of $0.3 M$ phosphorus acid $(H_{3} P O_{3})$ is:

When 100 ml of $\frac{M}{1 0} H_{2} S O_{4}$ is mixed with 500 ml of $\frac{M}{1 0}$ NaOH then nature of resulting solution and normality of excess of reactant left is

If 250 mL of a solution contains 24.5 g $H_{2} S O_{4}$ the molarity and normality respectively are:

10 mL N/2 HCl, 75 mL N/5 $H_{2} S O_{4}$ and 30 mL N/10 $H N O_{3}$ are mixed. Determine the normality of the mixture of the acids in solutions.

$100$ mL of $0.2 M$ $K A l (O H)_{2} C O_{3}$ solution is completely neutralized by a standard solution of $\frac{M}{4}$ $H_{2} C_{2} O_{4}$ . Which of the following is/are wrong?

What is the normality of 96% w/w aqueous $H_{2} S O_{4}$ solution?
(Atomic weight of $H = 1, S = 32, O = 16$ , density $= 1.84$ gm/L)

25 mL $H_{2} O_{2}$ were added to excess of acidified solution of KI. The iodine so liberated required 20 mL of 0.1 N sodium thiosulphate for titration. Calculate the volume strength.