Question

8. a. What is the molarity of a solution of calcium chloride which contains 33500 mg of solute in a total volume of 0.352L?

b. What is the molarity of a solution of LiI which contains 26800 mg of solute in a total volume of 0.282L?

c. What is the molarity of a solution of calcium chloride which contains 0.440moles of solute in a total volume of 0.872L?

9. a. How many mL of solvent should be added to 0.99L of a 1.5M solution if you want a final concentration of 0.74M.

b. 0.096L of a 2.9M solution is diluted to a final volume of 250mL. What is the
resultant molarity?
c. How many L of solvent should be added to 0.77L of a 1.0M solution if you want a
final concentration of 0.45M?

Answer 1

8. (a) As we know Molarity= No. of moles per litre

From question Given mass of solute is 33500 mg= 33.5 g

and Total volume is 0.352 L

We know molecular mass of sodium chloride is 110.98 g/mol

No. of moles =( Given mass in gram/ Molecular mass) = 33.5g/110.98 gmol^-1 = 0.3018 mole

Molecularity = No. of moles per litre= (0.3018/0.352) = 0.857 M

(b) As we know Molarity= No. of moles per litre

From question Given mass of solute is 26800 mg= 26.8 g

and Total volume is 0.282 L

We know molecular mass of Lithium iodide is 133.85 g/mol

No. of moles =( Given mass in gram/ Molecular mass) = (26.8/133.85) =0.2002 mol

So Molarity = No. of moles per litre = 0.2002/ 0.282 =0.71 M

(c) From question

Moles of solute = 0.44 mole

Total Volume = 0.872 L

So Molarity = No.of moles per litre = 0.44/0.872 =0.504 M

9. (a)

From Question

Molarity of solution = 1.5 M = 1.5 moles/ L

Volume of solution = 0.99 L

So No. of moles of solution = 1.5moles L^-1 * 0.99 L = 1.485 mole

From question

Final Concentration = 0.74 M

We know Molarity = No. of moles / Litre

Litre = No.of moles/ Molarity = 1.485 Moles/ 0.74 M = 2.0067 L

(b) From question

Molarity of solution = 2.9 M= 2.9 Moles L^-1

Volume of Solution = 0.096 L

So moles= 2.9 Moles L^-1 * 0.096 L = 0.2784 Mole

From question

Final Volume = 250 ml = 0.25 L

Resultant Molarity = 0.2784 Mole/ 0.25 L = 1.11 M

(c) From Question

Molarity of solution = 1.0 M = 1.0 moles/ L

Volume of solution = 0.77 L

So No. of moles of solution = 1.0moles L^-1 * 0.77 L = 0.77 mole

From question

Final Concentration = 0.45 M

We know Molarity = No. of moles / Litre

Litre = No.of moles/ Molarity = 0.77 Moles/ 0.45 M = 1.71 L