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QJEEAdv10Ch7371: JEE Advanced 2010 Chemistry - Class 11 JEEAdv10Ch7371 Solution: Silver (atomic weight = 108 g mol -1 ) has a density of 10.5…

Silver (atomic weight = 108 g mol-1) has a density of 10.5 g.cm-3. The number of silver atoms on a surface of area 10-12 m2 can be expressed in scientific notation as y ×\times 10x. The value of x is?

Step-by-Step Verified Solution

Given, atomic weight =108 g mol−1=108 \mathrm{~g} \mathrm{~mol}^{-1}

Density =10.5 g cm−3=10.5 \mathrm{~g} \mathrm{~cm}^{-3}

Surface area =10−12 m2=10^{-12} \mathrm{~m}^2

Volume of one silver atom =4/3πr3=4 / 3 \pi r^3

∵ Density = Mass  Volume ⇒ Volume = Mass  Density \because \text { Density }=\frac{\text { Mass }}{\text { Volume }} \Rightarrow \text { Volume }=\frac{\text { Mass }}{\text { Density }}

amp; or 43πr3=1086.023×1023×10.5amp;r3=108×36.023×1023×10.5×4×3.14amp;r3=0.40×10−23=4×10−24amp; or r=1.58×10−8 cm\begin{aligned}& \text { or } \frac{4}{3} \pi r^3=\frac{108}{6.023 \times 10^{23} \times 10.5} \\\\& r^3=\frac{108 \times 3}{6.023 \times 10^{23} \times 10.5 \times 4 \times 3.14} \\\\& r^3=0.40 \times 10^{-23}=4 \times 10^{-24} \\\\& \text { or } r=1.58 \times 10^{-8} \mathrm{~cm}\end{aligned}

No. of silver atoms on a surface area of 10−12 m210^{-12} \mathrm{~m}^2 can be given by 10−12=pr2×n10^{-12}=\mathrm{p} r^2 \times n

amp;n=10−123.14×(1.58×10−10)2=0.127×108⇒amp;n=1.27×107 or x=7\begin{aligned} & n=\frac{10^{-12}}{3.14 \times\left(1.58 \times 10^{-10}\right)^2}=0.127 \times 10^8 \\\\ \Rightarrow & n=1.27 \times 10^7 \text { or } x=7\end{aligned}

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