Question

We know that $S{O_2}$ turns limewater milky, so 3.2 g of $S{O_2}$ gives how much milkyness?

Answer 1

Hint – First write the equation for the reaction of $S{O_2}$ with limewater, i.e., $Ca{(OH)_2} + S{O_2} \to CaS{O_3} + {H_2}O$ , now 64 g of sulphur dioxide gives 120 g of milkyness, so using this find the answer.

Complete step-by-step answer:
We have been given that $S{O_2}$ turns limewater milky.
To find – 3.2 g of $S{O_2}$ will give milkyness.
Now, we know the molecular weight of $S{O_2}$ is $32 + 2 \times 16 = 64g$
Also, the molecular formula of lime water is $Ca{(OH)_2}$
So, the equation of reaction of sulphur dioxide with limewater is-
$Ca{(OH)_2} + S{O_2} \to CaS{O_3} + {H_2}O$
now, milkyness is due to the formation of $CaS{O_3}$
so, the molecular weight of $CaS{O_3}$ is $40 + 32 + 3 \times 16 = 120g$
Therefore, we can say that 64 g of sulphur dioxide gives 120 gm of milkyness.
So, we can write as 64 g of $S{O_2}$ gives 120 g milkyness
Therefore, 3.2 g of $S{O_2}$ will give milkyness $\dfrac{{120}}{{64}} \times 3.2 = 6g$
Hence, 3.2 g of sulphur dioxide gives 6 g of milkyness.

Note – Whenever such types of questions appear, then first write the things given in the question and then as mentioned in the solution, first we write the equation showing the reaction of sulphur dioxide with limewater and then we found out the molecular weight of sulphur dioxide and we have seen that 64 g of sulphur dioxide forms 120 g of milkyness, then we have found out that 3.2 g of sulphur dioxide forms 6 g of milkyness, which is the answer.