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 g2 g10 g20 g

Question

Toppr Admin · Accepted Answer

The correct option is A 1 gEquivalent weight of dibasic acid-xA0-mol-wt-2-2002-100Strength - 0-1 N- m - - V- 100 ml-From- m-E-xA0-V-xA0-N1000-100-xD7-100-xD7-0-11000-1g-xA0-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

The correct option is A 1 g
Equivalent weight of dibasic acid
$= \frac{m o l, w t .}{2} = \frac{200}{2} = 100$

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

From, $m = \frac{E V N}{1000} = \frac{100 \times 100 \times 0.1}{1000} = 1 g$

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 g2 g10 g20 g

The correct option is A 1 gEquivalent weight of dibasic acid =mol,wt.2=2002=100Strength = 0.1 N, m = ?, V= 100 ml.From, m=E V N1000=100×100×0.11000=1g 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

The correct option is A 1 g
Equivalent weight of dibasic acid
$= \frac{m o l, w t .}{2} = \frac{200}{2} = 100$

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

From, $m = \frac{E V N}{1000} = \frac{100 \times 100 \times 0.1}{1000} = 1 g$

Hence, the correct option is $A$