The number of moles of Ba(OH)_2=20×342100× 171=0.4 Moles of OH^ ions=0.4× 2=0.8 The number of moles of HNO_3=1.2× 2=2.4 Hence, the final solution is acidic due ...

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Standard XII

Chemistry

Molarity

342 g of 20

Question

$342 g$ of 20% (by mass) of $B a {(O H)}_{2}$ solution (specific gravity = 0.57) reacts with $1200 mL$ of $2 M$ $H N O_{3}$ . If the resulting density is the same as pure water, then the molarity of the hydronium ions in the resulting solution is:

0.25 M

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0.5 M

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0.888 M

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Solution

The correct option is C 0.888 M
The number of moles of $B a (O H)_{2} = \frac{20 \times 342}{100 \times 171} = 0.4$
Moles of $O H^{-}$ ions $= 0.4 \times 2 = 0.8$
The number of moles of $H N O_{3} = 1.2 \times 2 = 2.4$
Hence, the final solution is acidic due to the 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:

Q. 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 :

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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342g of 20% (by mass) of Ba(OH)2 solution (specific gravity = 0.57) reacts with 1200 mL of 2 M HNO3. If the resulting density is the same as pure water, then the molarity of the hydronium ions in the resulting solution is:

The correct option is C 0.888 MThe number of moles of Ba(OH)2=20×342100×171=0.4 Moles of OH− ions=0.4×2=0.8 The number of moles of HNO3=1.2×2=2.4 Hence, the final solution is acidic due to the presence of excess H+ H+=(2.4−0.8)×10001200+3420.57=0.888M

$342 g$ of 20% (by mass) of $B a {(O H)}_{2}$ solution (specific gravity = 0.57) reacts with $1200 mL$ of $2 M$ $H N O_{3}$ . If the resulting density is the same as pure water, then the molarity of the hydronium ions in the resulting solution is: