Question

1.A 693- mL sample of unknown HCl solution reacts completely with N a 2 C O 3 to form 20.1 g C O 2 . What was the concentration of the HCl solution?

2.The freezing point of a CaCl2 solution is -22 ∘C . What is the molality (m ) of the CaCl2 solution? What is the boiling point of the CaCl2 solution?

3. Calculate the molarity of 0.850 mole glucose in 0.250 L of a glucose solution.

Answer 1

Lets do it part by part.

1) Answer of this part is 1.3183 M.

Solution-

2) This is a question of freezing point depression.

The equation, we are going to use is -

ΔT_f = −i × K_f × m

T_f is the change in freezing point of the solvent,
        K_f is the molal freezing point depression constant, and

        m is the molal concentration of the solute in the solution

i = the number of dissolved particles (Van't Hoff Factor).

Now, K_f for water is 1.86°C/M.

i for CaCl₂ is 3 as CaCl₂ dissociates into 3 ions ( one Ca⁺² and 2 Cl^- )

ΔT_{f =} -22^o C

putting all the values, we get-

-22 = 3* 1.86 *m

so, m = 3.94 m

Similarly for boiling point, we will use-

ΔT = iK_bm

K_b water = 0.51 °C kg/mol
putting the values, we get-

ΔT = 3* 0.51 * 3.94 = 6.0282

So, the boiling point of the solution will be = 100+ 6.0282 = 106.0282 ^oC

3) Molarity = No of moles / Volume of the solution

No of moles of glucose = 0.850 (given)

Volume of solution = 0.250 L (given)

putting in the equation-

Molarity of glucose solution = 0.850 / 0.250 = 3.4 M

So, the answer = 3.4 M