2.56 × 10-3 equivalent of KOH is required to neutralise 0.12544 g H 2 XO 4. The atomic mass of X (in g / mol ) is: [Given: H 2 X O 4 is a dibasic acid]

Question

2.56 × 10-3 equivalent of KOH is required to neutralise 0.12544 g H 2 XO 4. The atomic mass of X (in g / mol ) is: [Given: H 2 X O 4 is a dibasic acid] (A) 16 (B) 8 (C) 7 (D) 32

Tardigrade Admin · Accepted Answer

Let the atomic weight of X = y No. of equivalents of KOH = No. of equivalents of H 2 XO 4 i.e. (0.12544/(66+y)2)=2.56× 10-3 ∴ y =32

Q. $2.56 \times 1 0^{- 3}$ equivalent of $K O H$ is required to neutralise $0.12544 g H_{2} X O_{4}$ . The atomic mass of $X$ (in $g / m o l$ ) is: [Given: $H_{2} X O_{4}$ is a dibasic acid]

$16$

$8$

$7$

$32$

Let the atomic weight of $X = y$
No. of equivalents of $K O H =$ No. of equivalents of $H_{2} X O_{4}$
i.e. $\frac{0.12544}{\frac{66 + y}{2}} = 2.56 \times 1 0^{- 3}$
$∴ y = 32$

Q. 2.56×10−3 equivalent of KOH is required to neutralise 0.12544gH2​XO4​. The atomic mass of X (in g/mol ) is: [Given: H2​XO4​ is a dibasic acid]

16

8

7

32

Let the atomic weight of X=y No. of equivalents of KOH= No. of equivalents of H2​XO4​ i.e. 266+y​0.12544​=2.56×10−3 ∴y=32

Q. $2.56 \times 1 0^{- 3}$ equivalent of $K O H$ is required to neutralise $0.12544 g H_{2} X O_{4}$ . The atomic mass of $X$ (in $g / m o l$ ) is: [Given: $H_{2} X O_{4}$ is a dibasic acid]

$16$

$8$

$7$

$32$

Let the atomic weight of $X = y$
No. of equivalents of $K O H =$ No. of equivalents of $H_{2} X O_{4}$
i.e. $\frac{0.12544}{\frac{66 + y}{2}} = 2.56 \times 1 0^{- 3}$
$∴ y = 32$