Question

) Calculate the osmolarity of each of the following solutions. i. 0.39 M Na2CO3 [5 marks] ii. 0.62 M Al(NO3)3 show working pls

Answer 1

Answer:

2.34

8.06

Explanation:

Osmolarity is calculated by determining the number of moles of a molecule into a solution and then multiplying the molarity with the number of particles produced by it

For the case of 0.39 M Na2CO3

Here Molarity is 0.39

Number of particles = 2 + 1 + 3 = 6

Osmolarity = 0.39 x 6 = 2.34

For the case of 0.62 M Al(NO3)3

Molarity = 0.62

Number of particles = 1 + (1+3) x 3 = 13

Osmolarity = 0.62 x 13 = 8.06

Answer 2

Given:

i) The molarity of Na2CO3 = 0.39 M

ii) The molarity of Al(NO3)3 = 0.62 M

To Find:

The osmolarity of each of the given solutions.

Calculation:

(i) Na2CO3 dissociates as follows:

Na2CO3 ⇄ 2 Na⁺ + CO3²⁻

⇒ n = 3

- As we know that osmolarity is the no of solute particles per litre solution, we have:

Osmolarity = n × M

⇒ Osmolarity = 3 × 0.39

⇒ Osmolarity = 1.17 Osm/L

(ii) Al(NO3)3 dissociates as:

Al(NO3)3 ⇄ Al³⁺ + 3 NO3⁻

⇒ n = 4

- As we know that osmolarity is the no of solute particles per litre solution, we have:

Osmolarity = n × M

⇒ Osmolarity = 4 × 0.62

⇒ Osmolarity = 2.48 Osm/L

- So, the osmolarity of Na2CO3 and Al(NO3)3 solutions is 1.17 Osm/L and 2.48 Osm/L respectively.