The mass of 98Molar Mass of S - 32 gmol-1- O - 16 gmol-1 and Na - 23 gmol-1

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Equivalent Mass

The mass of 9...

Question

The mass of 98% pure sulphuric acid required for the neutralization of 2 mol of $N a O H$ is:
(Molar Mass of S = 32 $g m o l^{- 1}$ , O = 16 $g m o l^{- 1}$ and Na = 23 $g m o l^{- 1}$ )

100 g

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98 g

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200 g

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196 g

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Solution

The correct option is A 100 g
$H_{2} S O_{4} + 2 N a O H \to N a_{2} S O_{4} + H_{2} O$

According to the stoichiometry, 2 moles of NaOH required 1 mole of sulphuric acid.
Mass of sulphuric acid required = 98 g
Mass of 98% sulphuric acid $\times \frac{98}{100} = 98$
$∴$ Mass of 98 % pure sulphuric acid required for the titration = 100 g

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Similar questions

Q. The mass of 98% pure sulphuric acid required for the neutralization of 2 mol of

N a O H

is:
(Molar Mass of S = 32

g m o l^{- 1}

, O = 16

g m o l^{- 1}

and Na = 23

g m o l^{- 1}

)

Q. The mass of 98% pure sulphuric acid required for the neutralization of 2 mol of

N a O H

is:
(Molar Mass of S = 32

g m o l^{- 1}

, O = 16

g m o l^{- 1}

and Na = 23

g m o l^{- 1}

)

Q. What weight of

80 %

pure

N a O H

required to neutralize

10 %

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100 m l, 1 M H_{2} S O_{4}

solution?
[Atomic mass of

N a = 23, H = 1, O = 16, S = 32

]

Q. An organic compound contains C,H and S the minimum molecular weight of the compound containing 10% sulphur is : (atomic weight of S=32 amu)

Pure $O_{2}$ diffuses through an aperture in 224 s, whereas a mixture of $O_{2}$ and another gas containing 80% $O_{2}$ diffuses from the same in 234s under similar conditions of pressure and temperature.What is molecular weight of gas?

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Equivalent Mass

Standard XIII Chemistry

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The mass of 98% pure sulphuric acid required for the neutralization of 2 mol of NaOH is: (Molar Mass of S = 32 gmol−1, O = 16 gmol−1 and Na = 23 gmol−1)

The correct option is A 100 gH2SO4+2NaOH→Na2SO4+H2O According to the stoichiometry, 2 moles of NaOH required 1 mole of sulphuric acid. Mass of sulphuric acid required = 98 g Mass of 98% sulphuric acid ×98100=98 ∴ Mass of 98 % pure sulphuric acid required for the titration = 100 g

The mass of 98% pure sulphuric acid required for the neutralization of 2 mol of $N a O H$ is:
(Molar Mass of S = 32 $g m o l^{- 1}$ , O = 16 $g m o l^{- 1}$ and Na = 23 $g m o l^{- 1}$ )