Find the weight of $N a N O_{3}$ in grams (Rounded off to nearest integer), required to make 50 mL of an aqueous solution containing 70 mg per mL of $N a^{+}$ ions

[Given:Atomic weight in $g m o l^{- 1} [N a : 23; N : 14; O : 16]$

 JEE MAIN 2021

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Solution

$Concentration of N a^{+} = 70 m g / m L$
$W_{N a}$ in 50 mL solution:
$70 \times 50 m g = 3500 m g = 3.5 g$
$Moles of N a^{+} in 50 mL solution = \frac{3.5}{23} m o l$
$N a N O_{3} (a q) \to N a^{+} (a q) + N O_{3}^{-} (a q)$
$Moles of N a N O_{3} = Moles of N a^{+} = \frac{3.5}{23} m o l$
$Mass of N a N O_{3} = \frac{3.5}{23} \times 85 = 12.94 \approx 13 g$

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Similar questions

Q. Find the weight of

N a N O_{3}

in grams (Rounded off to nearest integer), required to make 50 mL of an aqueous solution containing 70 mg per mL of

N a^{+}

ions

[Given:Atomic weight in

g m o l^{- 1} [N a : 23; N : 14; O : 16]

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N a N O_{3}

weighed to make

50 m L

of an aqueous solution containing 70.0

m g

N a^{+}

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is _____

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. (Rounded off to the nearest integer)

[Given: Atomic weight in

g m o l^{- 1}

. Na: 23; N: 14; O: 16.]

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Find the weight of NaNO3 in grams (Rounded off to nearest integer), required to make 50 mL of an aqueous solution containing 70 mg per mL of Na+ ions [Given:Atomic weight in gmol−1[Na:23;N:14;O:16]  JEE MAIN 2021

Concentration of Na+=70 mg/mL WNa in 50 mL solution: 70×50mg=3500 mg=3.5g Moles of Na+ in 50 mL solution =3.523 mol NaNO3 (aq)→Na+ (aq)+NO−3 (aq) Moles of NaNO3=Moles of Na+=3.523 mol Mass of NaNO3=3.523×85=12.94≈13g