Question

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

Answer 1

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. 

Answer 2

Answer : The new concentration will be, 0.0757 M

Solution :

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of NaCl solution = 0.455 M

$V_1$ = volume of NaCl solution = 375 ml = 0.375 L

$M_2$ = new concentration = ?

$V_2$ = volume of new concentration after dilution with water = 0.375 + 1.88 = 2.255 L

Now put all the given values in the above law, we get the new concentration.

$(0.455M)\times 0.375L=M_2\times (2.255L)$

$M_2=0.0757M$

Therefore, the new concentration will be, 0.0757 M