(B) Total mass of $Na^{+}$ required $= 70.0 \ mg/mL \times 50 \ mL = 3500 \ mg = 3.5 \ g$.Molar mass of $NaNO_3 = 23 + 14 + (3 \times 16) = 85 \ g \ m

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(B) Total mass of $Na^{+}$ required $= 70.0 \ mg/mL \times 50 \ mL = 3500 \ mg = 3.5 \ g$.Molar mass of $NaNO_3 = 23 + 14 + (3 \times 16) = 85 \ g \ mol^{-1}$.Moles of $Na^{+}$ $= \frac{3.5 \ g}{23 \ g \ mol^{-1}} \approx 0.15217 \ mol$.Since $1 \ mol$ of $NaNO_3$ contains $1 \ mol$ of $Na^{+}$,moles of $NaNO_3 = 0.15217 \ mol$.Mass of $NaNO_3 = 0.15217 \ mol \times 85 \ g \ mol^{-1} = 12.93445 \ g$.Rounding off to the nearest integer,we get $13 \ g$.

Class 11 Chemistry Some Basic Concepts of Chemistry Chemical stoichiometry

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The $NaNO_3$ weighed out to make $50 \ mL$ of an aqueous solution containing $70.0 \ mg$ $Na^{+}$ per $mL$ is $...... \ g$.
(Rounded off to the nearest integer) [Given : Atomic weight in $g \ mol^{-1}$ - $Na : 23$; $N : 14$; $O : 16$]

JEE MAIN 2021 All JEE MAIN

A
$19$
B
$13$
C
$17$
D
$10$

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Some Basic Concepts of Chemistry

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Chemical stoichiometry

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The $NaNO_3$ weighed out to make $50 \ mL$ of an aqueous solution containing $70.0 \ mg$ $Na^{+}$ per $mL$ is $...... \ g$.(Rounded off to the nearest integer) [Given : Atomic weight in $g \ mol^{-1}$ - $Na : 23$; $N : 14$; $O : 16$]

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The $NaNO_3$ weighed out to make $50 \ mL$ of an aqueous solution containing $70.0 \ mg$ $Na^{+}$ per $mL$ is $...... \ g$.
(Rounded off to the nearest integer) [Given : Atomic weight in $g \ mol^{-1}$ - $Na : 23$; $N : 14$; $O : 16$]