100 ml of ethyl alcohol [d = 0.92 g/ml] and 900 ml of water [d = 1 g/ml] are mixed to form 1 lit solution. The Molarity and molality of the resulting solution are

$\mathrm{Molarity}=\frac{\mathrm{no}.\mathrm{of} \mathrm{moles}}{\mathrm{volume}\left(\mathrm{l}\right)}$

Mass of ethyl alcohol 
Mass=density×volume ⇒Mass=100×0.92=92g
Moles of ethyl alcohol=$={\displaystyle \frac{92}{46}}=2$moles.
Thus, molarity=2M

So ${\displaystyle \frac{92}{900}}\times 1000$g of ethyl alcohol is present.
=102.22g
moles=102.22/46
Hence the answer is = 2.22

correct answer is (B) 2M and 2.22m