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 :
Volume of the gold dispersed in one litre water \[=\frac{\text{Mass}}{\text{Density}}\] \[=\frac{1.9\times {{10}^{-4}}\,gm}{19\ gm\ c{{m}^{-3}}}\] \[=1\times {{10}^{-5}}c{{m}^{-3}}\] Radius of gold sol particle \[=10\ nm\] \[=10\times {{10}^{-9}}m\] \[=10\times {{10}^{-7}}cm\ ={{10}^{-6}}cm\] Volume of the gold sol particle \[=\frac{4}{3}\pi {{r}^{3}}\] \[=\frac{4}{3}\times \frac{22}{7}\times {{({{10}^{-6}})}^{3}}\] \[=4.19\times {{10}^{-18}}c{{m}^{3}}\] No. of gold sol particle in \[1\times {{10}^{-5}}c{{m}^{3}}\]\[=\frac{1\times {{10}^{-5}}}{4.19\times {{10}^{-18}}}\] \[=2.38\times {{10}^{12}}\] No. of gold sol particle in one \[m{{m}^{3}}\] \[=\frac{2.38\times {{10}^{12}}}{{{10}^{6}}}=2.38\times {{10}^{6}}\]
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