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}} $