$100 \mathrm{~mL}$ of $\mathrm{N} / 5 \mathrm{NaOH}$ will neutralize

$1.2368 \mathrm{~g}$ of $\mathrm{H}_3 \mathrm{BO}_3$

$0.0618 \mathrm{~g}$ of $\mathrm{H}_3 \mathrm{BO}_3$

$0.1855 \mathrm{~g}$ of $\mathrm{H}_3 \mathrm{BO}_3$

$0.03092 \mathrm{~g}$ of $\mathrm{H}_3 \mathrm{BO}_3$

JEE CHEMISTRY MCQ: SOME BASIC CONCEPTS OF CHEMISTRY

1 Medium

1 of 1

Application +4 | -1

100 \mathrm{~mL}

\mathrm{N} / 5 \mathrm{NaOH}

will neutralize

View answer and explanation

Correct option: C

1.2368 \mathrm{~g}

\mathrm{H}_3 \mathrm{BO}_3

Option C (

1.2368 \text{{ g}}

\mathrm{H}_3 \mathrm{BO}_3

) is correct because it is calculated by equating the equivalents of NaOH and H3BO3, considering the n-factor of H3BO3 is 1.

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