7: If 120mL of a 6 MNaOH solution is diluted to a final volume of 600mL , what i

ID: 865320 • Letter: 7

Question

7: If 120mL of a 6 MNaOH solution is diluted to a final volume of 600mL , what is the resulting concentration of the solution?

8: What is the molarity of 0.50L of solution containing 46.0g of CaCO3?

9: If one tablet weighs 1.25g , what percentage of the tablet is the active ingredient?

10: a 49.0mL sample of 3.7{ m \%} (m/v) NaOH is diluted with water so that the final volume is 106.0mL .

11: a 15.0mL sample of 35{ m \%} (m/v) acetic acid (CH3COOH) solution is added to water to give final volume of 26mL .

7-11 please

7)Moles of solute = molarity*volume of solution in litres

Since, moles of NaOH remains constant

Thuerefore, Applying M1*V1 = M2*V2 ; we get

6*0.12 = M2*0.6

or, M2 = 1.2 M

Hence, final molar concentration = 1.2 M

8) Molar mass of CaCO3 = 100 g/mole

Thus, moles of CaCO3 in 46 g of it = mass/molar mass = 46/100 = 0.46 moles

Now, molarity = moles of CaCO3/Volume of solution in litres = 0.46/0.5 = 0.92 M

9) Question is incomplete

10) 3.7 mass/ volume % means 3.7 g of NaOH is present in 100 ml of solution

Thus, in 49 ml of the solution, mass of NaOH present = 3.7*49/100 = 1.813 g

Now, final volume of the solution = 106 ml

Thus, mass/volume % = (mass of NaOH/volume of solution)*!00 = (1.813/106)*100 = 1.7104 %

11) 35 mass/ volume % means 35 g of acetic acid is present in 100 ml of solution

Thus, in 15 ml of the solution, mass of NaOH present = 35*15/100 = 5.25 g

Now, final volume of the solution = 26 ml

Thus, mass/volume % = (mass of acetic acid/volume of solution)*!00 = (5.25/26)*100 = 20.192 %

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