The correct option is D 540 g
The molecular formula of urea =NH2CONH2
So, in 60 g of urea, mass of nitrogen =28 g.
% nitrogen = 2860×100=46.66%
The sample is impure.
According to the question, the mass of nitrogen in 560 g of the sample =560×45100=252 g.
So, amount of urea when 252 g of nitrogen is present =6028×252=540 g.