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Standard XI

Chemistry

Molarity

100 mL of H2S...

Question

$100 m L o f H_{2} S O_{4}$ solution having molarity $1 M$ and density $1.5 g / m L$ is mixed with $400 m L$ water. Calculate final molarity of $H_{2} S O_{4}$ solution, if final density is $1.25 g / m L$ :

4.4 M

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0.145 M

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0.52 M

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0.227 M

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Solution

The correct option is D $0.227 M$
$Total moles of H_{2} S O_{4} = 0.1 m o l e$
$Total volume = \frac{150 + 400}{1.25} = \frac{550}{1.25} = 440$
$∴ M = \frac{0.1}{440} \times 1000 = \frac{1}{4.4} = 0.227 M$

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Similar questions

100 m L o f H_{2} S O_{4}

solution having molarity

1 M

and density

1.5 g / m L

is mixed with

400 m L

water. Calculate final molarity of

H_{2} S O_{4}

solution, if final density is

1.25 g / m L

100 mL

of an

H_{2} S O_{4}

solution that has a molarity of

1 M

and density of

1.50 g/mL

is mixed with

400 mL

of water. If the final density of the resulting solution is

1.25 g/mL

, the final molarity of the

H_{2} S O_{4}

solution is:

100 mL

of an

H_{2} S O_{4}

solution that has a molarity of

1 M

and density of

1.50 g/mL

is mixed with

400 mL

of water. If the final density of the resulting solution is

1.25 g/mL

, the final molarity of the

H_{2} S O_{4}

solution is:

100 mL

H C l

solution having molarity

1 M

is mixed with

400 mL

of water. Calculate final molarity of

H C l

solution.

Q. A solution is made by mixing equal volumes of

30 %

by weight of

H_{2} S O_{4}

(density =

1.218

g/mL) and

70 %

by weight of

H_{2} S O_{4}

(density =

1.610

g/mL). Then find the molarity of the final solution?

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100 mL of H2SO4 solution having molarity 1 M and density 1.5 g/mL is mixed with 400 mL water. Calculate final molarity of H2SO4 solution, if final density is 1.25 g/mL:

The correct option is D 0.227 MTotal moles of H2SO4=0.1 mole Total volume=150+4001.25=5501.25=440 ∴M=0.1440×1000=14.4=0.227 M

$100 m L o f H_{2} S O_{4}$ solution having molarity $1 M$ and density $1.5 g / m L$ is mixed with $400 m L$ water. Calculate final molarity of $H_{2} S O_{4}$ solution, if final density is $1.25 g / m L$ :

The correct option is D $0.227 M$
$Total moles of H_{2} S O_{4} = 0.1 m o l e$
$Total volume = \frac{150 + 400}{1.25} = \frac{550}{1.25} = 440$
$∴ M = \frac{0.1}{440} \times 1000 = \frac{1}{4.4} = 0.227 M$