Question

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

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

M_dilution V_dilution = M_stock V_stock

M_dilution x 100mL = 1.0 M x 10 mL

M_dilution = 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

M_dilution V_dilution = Mx solution Vx solution

M_dilution x 95 mL = 0.1 M x 5.0 mL

M_dilution = 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