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Standard X

Chemistry

Relationship between Vapor Density & Relative Molecular Mass

342 gm of 2...

Question

342 gm of $20 %$ by mass of $B a (O H)_{2}$ solution (sp.gr.0.57) is reacted with 1200 mL of 2 M $H N O_{3}$ .If the final density is same as pure water than molarity of the ion in resulting solution by nature of the above solution is identified, is :

0.25

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0.5 M

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0.888 M

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Solution

The correct option is D 0.888 M
No. of moles of $B a (O H)_{2} = \frac{20 \times 342}{100 \times 171}$
$= 0.4$ mole
or moles of $O H^{-} = 0.4 \times 2$
No. of moles of $H N O_{3} = 1.2 \times 2 = 2.4$ mole
Hence, the final solution is acidic due to presence of excess $H^{+}$
$[H^{+}] = \frac{(2.4 - 0.8) \times 1000}{1200 + \frac{342}{0.57}} = 0.888 M$

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Similar questions

342 g

of 20% (by mass) of

B a {(O H)}_{2}

solution (specific gravity = 0.57) reacts with

1200 mL

2 M

H N O_{3}

. If the resulting density is the same as pure water, then the molarity of the hydronium ions in the resulting solution is:

342 g

of 20% (by mass) of

B a {(O H)}_{2}

solution (specific gravity = 0.57) reacts with

1200 mL

2 M

H N O_{3}

. If the resulting density is the same as pure water, then the molarity of the hydronium ions in the resulting solution is:

342

g of

20 %

by mass of

B a (O H)_{2}

solution (specific gravity

= 0.57

) is reacted with

200

mL of

2 M H N O_{3}

according to given balanced reaction.

B a (O H)_{2} + 2 H N O_{3} ⟶ B a (N O_{3})_{2} + 2 H_{2} O

The nature of the final 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

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:

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342 gm of 20% by mass of Ba(OH)2 solution (sp.gr.0.57) is reacted with 1200 mL of 2 M HNO3.If the final density is same as pure water than molarity of the ion in resulting solution by nature of the above solution is identified, is :

The correct option is D 0.888 MNo. of moles of Ba(OH)2=20×342100×171 =0.4 mole or moles of OH−=0.4×2 No. of moles of HNO3=1.2×2=2.4 mole Hence, the final solution is acidic due to presence of excess H+ [H+]=(2.4−0.8)×10001200+3420.57=0.888M

342 gm of $20 %$ by mass of $B a (O H)_{2}$ solution (sp.gr.0.57) is reacted with 1200 mL of 2 M $H N O_{3}$ .If the final density is same as pure water than molarity of the ion in resulting solution by nature of the above solution is identified, is :