(C) The neutralization reaction is: $2HNO_3 + Ba(OH)_2 \rightarrow Ba(NO_3)_2 + 2H_2O$. According to the law of equivalence,the number of equivalents

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(C) The neutralization reaction is: $2HNO_3 + Ba(OH)_2 \rightarrow Ba(NO_3)_2 + 2H_2O$. According to the law of equivalence,the number of equivalents of $HNO_3$ must equal the number of equivalents of $Ba(OH)_2$. $n_{factor} \times M_1 \times V_1 = n_{factor} \times M_2 \times V_2$. For $HNO_3$,$n_{factor} = 1$. For $Ba(OH)_2$,$n_{factor} = 2$. $1 \times 0.04 \times V_1 = 2 \times 0.01 \times 50$. $0.04 \times V_1 = 1$. $V_1 = \frac{1}{0.04} = 25 \ mL$.

Class 11 Chemistry Some Basic Concepts of Chemistry Chemical stoichiometry

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The volume of $0.04 \ M$ aqueous solution of $HNO_3$ required to neutralize $50 \ mL$ of $0.01 \ M$ aqueous solution of $Ba(OH)_2$ is (Assume complete neutralization). (in $mL$)

A
$15$
B
$20$
C
$25$
D
$40$

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The volume of $0.04 \ M$ aqueous solution of $HNO_3$ required to neutralize $50 \ mL$ of $0.01 \ M$ aqueous solution of $Ba(OH)_2$ is (Assume complete neutralization). (in $mL$)

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The molarity of $HNO_{3}$ in a sample which has density $1.4 \; g/mL$ and mass percentage of $63\%$ is . . . . . . $M$.
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