Solutions Question 246
Question: The molality of a urea solution in which 0.0200 g of urea ( $ NH_2CONH_2 $ ) is added to 0.400 $ dm^{3} $ of water at STP is
Options:
A) $ 0.555,mol,k{g^{-1}} $
B) $ 5.55\times {10^{-4}},mol,k{g^{-1}} $
C) $ 8.33\times {10^{-4}},mol,k{g^{-1}} $
D) $ 33.3,mol,k{g^{-1}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ 1,dm^{3}=100,cm^{3}=1,L=1,kg $
$ 0.400,d{m^{^{3}}} $ of $ H_2O $ at STP= $ 0.400,kg $ of $ H_2O $ (Density=1.0 g $ c{m^{-3}} $ ) Moles of urea = $ \frac{0.0200}{60}mol $ (Molar mass of $ NH_2,CONH_2=60gmo{l^{-1}} $ )
$ \therefore $ Molality $ =\frac{Moles,of,solut,\text{(urea)}}{kg,of,solvent}=\frac{0.020/60}{0.400kg} $
$ =8.33\times {10^{-4}}mol,k{g^{-1}} $
