2 g of 109 So, gm eq. of H_2SO_4= 2.18/98× 2 Similarly, gm eq. of Ca(OH)_2=44.49 × 10^ 3× M × 2 Since, gm eq. of H_2SO_4=gm eq. of Ca(OH)_2 Therefore, 2.18/98× ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Introduction to Oxidation and Reduction

2 g of 109

Question

2 g of 109% oleum is diluted with water. To neutralize this solution completely, 44.49 ml of $C a (O H)_{2}$ solution is required. What is the molarity of $C a (O H)_{2}$ solution?

1.0 M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.1 M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.05 M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.5 M

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 0.5 M
2 g of 109% oleum $= 2 \times 1.09 = 2.18$ g of $H_{2} S O_{4}$ in dilute solution.
So, gm eq. of $H_{2} S O_{4} = \frac{2.18}{98} \times 2$
Similarly, gm eq. of $C a (O H)_{2} = 44.49 \times 10^{- 3} \times M \times 2$
Since,
$g m e q . o f H_{2} S O_{4} = g m e q . o f C a (O H)_{2}$
Therefore,
$\frac{2.18}{98} \times 2 = 44.49 \times 10^{- 3} \times M \times 2$
$\Rightarrow M = 0.5$

Suggest Corrections

Similar questions

Q. 2 g of 109% oleum is diluted with water. To neutralize this solution completely, 44.49 ml of

C a (O H)_{2}

solution is required. What is the molarity of

C a (O H)_{2}

solution?

Q. 5 g of oleum is diluted with water. It requires 533.16 ml of 0.1 M

B a (O H)_{2}

solution for complete neutralization. Find oleum strength (in %).

Q. If

50 m L

0.1 M C a (O H)_{2}

solution is required to neutralize

25 m L

H_{3} P O_{4}

solution then what will be the strength of

H_{3} P O_{4}

g / L

1

g of oleum sample is diluted with water. The solution required

54

mL of

0.4 N

N a O H

for complete neutralization. The percentage of free

{S O}_{3}

in the sample is :

0.5

g of fuming

H_{2} S O_{4}

(oleum) is diluted with water. The solution requires

26.7

mL of

0.4 N N a O H

for complete neutralization. Find the percentage of free

S O_{3}

in the sample of oleum.

Join BYJU'S Learning Program

2 g of 109% oleum is diluted with water. To neutralize this solution completely, 44.49 ml of Ca(OH)2 solution is required. What is the molarity of Ca(OH)2 solution?

The correct option is D 0.5 M2 g of 109% oleum =2×1.09=2.18 g of H2SO4 in dilute solution. So, gm eq. of H2SO4=2.1898×2 Similarly, gm eq. of Ca(OH)2=44.49×10−3×M×2 Since, gm eq. of H2SO4=gm eq. of Ca(OH)2 Therefore, 2.1898×2=44.49×10−3×M×2 ⇒M=0.5

2 g of 109% oleum is diluted with water. To neutralize this solution completely, 44.49 ml of $C a (O H)_{2}$ solution is required. What is the molarity of $C a (O H)_{2}$ solution?