Neet Solved Paper 2013 Question 39
Question: $ 6.02\times 10^{20} $ molecules of urea are present in 100 mL of its solution. The concentration of solution is
Options:
A) 0.02 M
B) 0.01 M
C) 0.001 M
D) 0.1 M
Show Answer
Answer:
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}} $
