A solution X is made by taking 10.00 ml of a 1.0 M solution and add enough water to make 100.0 ml total volume. Solution Y is made by taking 5.0 ml of solution X and diluting it again by adding it to 95 ml of water. Calculate the molarity of solutions X and Y.
Calculate the molarity of the solution created when 50.0 ml of 0.132 M NaOH is mixed with 75.0 ml of 0.120 M NaOH ( assuming that the volumes are additive)
ANSWER 1
Molarity of X solution
We have initial solution 10 mL (i.e. 10/1000 = 0.01 L) of molarity 1.0 M
Volume of stock solution = 10 mL
Molarity of stock solution = 1.0 M
Volume of dilution = 100 mL
Here we know the
Mdilution Vdilution = Mstock Vstock
Mdilution x 100mL = 1.0 M x 10 mL
Mdilution = 0.1 M
Molarity of X solution= 0.1 M
ANSWER 2
Similarly
Molarity of Y solution
Calculated molarity x solution 0.1 M and volume of x solution used to made Y = 5.0 mL
Volume of x solution = 5.0 mL
Molarity of x solution = 0.1 M
Volume of dilution = 95 mL
Here we know the
Mdilution Vdilution = Mx solution Vx solution
Mdilution x 95 mL = 0.1 M x 5.0 mL
Mdilution = 0.0053M
Molarity of Y solution= 0.0053M
ANSWER 3
Now molarity of NaOH solution
We have two solution one is 50.0 mL (i.e. 50 mL = 0.050 L) of 0.132 M NaOH
Number of moles of NaOH = 0.132 mol/ L x 0.050 L
Number of moles of NaOH = 0.0066 mole
Similarly for second solution 75.0 mL (i.e. 75 mL = 0.075 L) of 0.120 M NaOH
Number of moles of NaOH = 0.120 mol/ L x 0.075 L
Number of moles of NaOH = 0.009 mole
Total number of moles of NaOH = 0.0066 + 0.009 = 0.0156 mole
Total volume of solution = 0.050 +0.075 = 0.125 L
Now molarity of NOH solution = number mole / volume in L
Now molarity of NOH solution = 0.0156 moles / 0.125 L =0.1248 M
Molarity of NaOH solution after mixing of two solution =0.125 M
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