A) \[1.9\times {{10}^{12}}\]
B) \[6.3\times {{10}^{14}}\]
C) \[6.3\times {{10}^{10}}\]
D) \[2.4\times {{10}^{6}}\]
Correct Answer: D
Solution :
[d] Volume of gold dispersed in 1 L water \[=\frac{Mass}{Density}=\frac{1.9\times {{10}^{-4}}}{19gm\,c{{m}^{-3}}}=1\times {{10}^{-5}}c{{m}^{3}}\] Radius of gold sol particle \[=10nm=10\times {{10}^{-7}}cm={{10}^{-6}}cm\] Volume of gol sol particle \[=\frac{4}{3}\pi {{r}^{3}}\] \[=\frac{4}{3}\times \frac{22}{7}\times {{({{10}^{-6}})}^{3}}=4.19\times {{10}^{-18}}c{{m}^{3}}\] \[\therefore \] No. of gold sol particles in \[1\times {{10}^{-5}}c{{m}^{3}}\] \[=\frac{1\times {{10}^{-5}}}{4.19\times {{10}^{-18}}}=2.38\times {{10}^{12}}\] \[\therefore \] No. of gold sol particles in one \[m{{m}^{3}}\] \[=\frac{2.38\times {{10}^{12}}}{{{10}^{6}}}=2.38\times {{10}^{6}}\]
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