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JEE Advanced 2010Paper 2
Exam
JEE Advanced
Year
2010
Paper
Paper 2
Subject
Chemistry
Chapter
Solid State

JEE Advanced 2010 · Paper 2 · Chemistry · Solid State

Silver (atomic weight =108 ~g ~mol^ -1 ) has a density of 10.5 ~g...

Silver (atomic weight =108 g mol−1=108 \mathrm{~g} \mathrm{~mol}^{-1} ) has a density of 10.5 g cm−310.5 \mathrm{~g} \mathrm{~cm}^{-3}. The number of silver atoms on a surface of area 10−12 m210^{-12} \mathrm{~m}^2 can be expressed in scientific notation as y×10xy \times 10^x. The value of xx is

Correct answer

0

Step-by-step solution

Volume of one mole of silver atoms =10810.5cm3\frac{108}{10.5} \mathrm{ cm}^3 /mole Volume of onesilveratom=10810.5×16.022×1023cm3So,one silver atom =\frac{108}{10.5} \times \frac{1}{6.022 \times 10^{23}} \mathrm{ cm}^3 So,43πR3=10810.5×16.022×1023=1.708×10−23R3=0.407×10−23cm3=0.407×10−29m3\begin{aligned} \frac{4}{3} \pi R^3 & =\frac{108}{10.5} \times \frac{1}{6.022 \times 10^{23}}=1.708 \times 10^{-23} \\ R^3 & =0.407 \times 10^{-23} \mathrm{ cm}^3 \\ & =0.407 \times 10^{-29} \mathrm{ m}^3 \end{aligned}AreaofeachsilveratomπR2=π×(0.407×10−29m3)2/3Sonumberofsilveratomsingivenarea=10−12(0.407×10−29m3)2/3=108(π×2)=1.6×107=y×10x so, x=7SolidstateMathematicalunderstandingIIIArea of each silver atom \pi R^2=\pi \times\left(0.407 \times 10^{-29} \mathrm{ m}^3\right)^{2 / 3} So number of silver atoms in given area =\frac{10^{-12}}{\left(0.407 \times 10^{-29} \mathrm{ m}^3\right)^{2 / 3}}=\frac{10^8}{(\pi \times 2)}=1.6 \times 10^7=y \times 10^x \quad \text{ so, } \quad x=7 Solid state Mathematical understanding III
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