A crystalline hydrated salt on being rendered anhydrous loses 45.6% of its weight.
The percentage composition of the anhydrous salt is:
$A l = 10.5 %, K = 15.1 %, S = 24.8 %$ and oxygen $= 49.6 %$
[Molar mass of $A l = 27, K = 39, S = 32$ ].

What is the empirical formula of the hydrated salt?

K A l S_{2} O_{8} .10 H_{2} O

No worries! We‘ve got your back. Try BYJU‘S free classes today!

K A l S_{2} O_{8} .16 H_{2} O

No worries! We‘ve got your back. Try BYJU‘S free classes today!

K A l S_{2} O_{8} .8 H_{2} O

No worries! We‘ve got your back. Try BYJU‘S free classes today!

K A l S_{2} O_{8} .12 H_{2} O

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $K A l S_{2} O_{8} .12 H_{2} O$
Assume molecular formula of hydrated salt = $K A l S_{2} O_{8} . x H_{2} O$

Then, $\frac{18 x}{39 + 27 + 64 + 128 + 18 x} \times 100 = 45.6$
$\Rightarrow x = 12$ .

Suggest Corrections

Similar questions

45.6 %

A l - 10.5 %, K - 15.1 %, S - 24.8 %, O - 49.6 % .

M = 948 {g mol}^{- 1}

A l =

K =

S =

A l =

K =

S =

O =

Join BYJU'S Learning Program