375ml of a 0.455m sodium chloride solution is dilluted with 1.88l of water. what is the new concentration in molarity

0.375 L of the NaCl solution contains 0.455 moles  (1)

            since it was diluted to 1.88 L

then let   1.88 L of solution contain  x  moles  (2)

            by combining both (1) and (2)      
       
        ⇒     (0.345 L) x = (0.455 mol) (1.88 L)

       ⇒                     x = (0.242 mol · L)  ÷ (0.345 L)

       ⇒                     x =  0.702 mol

                
Thus the molarity of the NaCl after being diluted is 0.702 M  OR 0.702 mol / L   OR  0.702 mol/dm³. 

The mole ratio of hydrogen to ammonia is 3: 2.