Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

3.01e22 Ag+ ions is present in? options a) 85 grams of AgNO3 b) 0.85g AgNO3 c) 8.5g AgNO3 d) 18.5g AgNO3

1 Answer

Sarita Rana · Nam D. · Soumalya Pramanik

Feb 2, 2018

$8.5$ $8.5$ g of $A g N O_{3}$ $AgNO_3$

Explanation:

$1$ $1$ mole $= 6.023 \times 10^{23}$ $= 6.023 xx 10^23$ ions
$A g^{+} =$ $Ag^+ =$ $3.01 \times 10^{22}$ $3.01 xx 10^22$ ions

No. of moles of $A g^{+} =$ $Ag^+ =$ ( $3.01$ $3.01$ x $10^{22}$ $10^22$ ions)/( $6.023 \times 10^{23}$ $6.023 xx 10^23$ ions) $=$ $=$

$0.0499$ $0.0499$ moles

Molar mass of $A g N O_{3}$ $AgNO_3$ $=$ $=$ $169.87 g / m o l$ $169.87 g"/"mol$

Mass of $A g N O_{3}$ $AgNO_3$ $=$ $=$ $0.0499$ $0.0499$ moles $\times 169.87 g / m o l = 8.47 g$ $xx 169.87 g"/"mol = 8.47 g$

$AgNO_3 -> Ag^+ \ + \ NO_3^-$

Related questions

Impact of this question

2687 views around the world

You can reuse this answer
Creative Commons License