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Question

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A. 2

B. 4

C. 6

D. 8

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-To find the value of atomicity we will assume it as 'n'. So, the molecular formula will become \[{{\text{S}}_{n}}\].

-Now, we all know that the molecular mass of the sulphur is 32 so the mass of the compound of sulphur will be 32n.

-Here, 32 is the mass of sulphur and n is the mass of an atom whose value we have to find.

-Now, let's apply the formula depression in freezing point i.e.

$\Delta {{\text{T}}_{f}}\text{ = }\dfrac{100\text{ w }{{\text{K}}_{f}}}{\text{mW}}$

-Here, \[{{\text{K}}_{f}}\] expresses the molal freezing point constant, $\Delta {{\text{T}}_{f}}$ is the depression in the freezing point or the difference between the freezing point of pure solvent and the freezing point of the solution.

-Whereas w is the weight of the solute in gram, W is the weight of solvent in kg and m is the molecular weight of solute.

-It is given that the value of depression in freezing point is $\text{0}\text{.010}{}^\circ \text{C}$, w is the mass of sulphur i.e.2.56g and w is the mass of compound 100g. We have to find the value m that is atomicity.

-Now, by applying the formula of depression in freezing point we will get:

$0.010\ \text{= }\dfrac{1000\ \times \text{ 2}\text{.56 }\times \text{ 0}\text{.1}}{\text{m }\times \text{ 100}}$

$\text{m}\ \text{= }\dfrac{1000\ \times \text{ 2}\text{.56 }\times \text{ 0}\text{.1}}{\text{0}\text{.010 }\times \text{ 100}}$

$32\text{n = 256}$

\[\text{n = }\dfrac{\text{256}}{32}\text{ = 8}\]

-Therefore, the atomicity of sulphur is 8 and the molecular formula will be \[{{\text{S}}_{8}}\].