A) 0.02 M
B) 0.01 M
C) 0.001 M
D) 0.1 M
Correct Answer: B
Solution :
Given, number of molecules of urea \[=6.02\times {{10}^{20}}\] \[\therefore \]Number of moles \[=\frac{6.02\times {{10}^{20}}}{{{N}_{A}}}\] \[=\frac{6.02\times {{10}^{20}}}{6.02\times {{10}^{23}}}=1\times {{10}^{-3}}mol\]Volume of the solution \[=100\,mL=\frac{100}{1000}L=0.1L\] Concentration of urea solution (in mol \[{{L}^{-1}}\])\[=\frac{1\times {{10}^{-3}}}{0.1}=mol\,{{L}^{-1}}\] \[=1\times {{10}^{-2}}mol\,{{L}^{-1}}=0.01\,mol\,{{L}^{-1}}\]
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