Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

4.74 g of an inorganic compound contains 0.39 g of K, 0.27 g of Al, 1.92 g of $SO_4$ radicals and 2.16 g of water. If molar mass of the compound is $948 \text{ g mol}^{-1}$ , the molecular formula of the inorganic compound is:

KAl(SO₄)₂·12H₂O

K₂Al₂(SO₄)₆·12H₂O

K₂SO₄·Al₂(SO₄)₃·24H₂O

K₂SO₆·Al₂(SO₄)₃·12H₂O

Step-by-Step Solution

Calculate Moles of Constituents:

Moles of K = $\frac{0.39}{39} = 0.01 \text{ mol}$
Moles of Al = $\frac{0.27}{27} = 0.01 \text{ mol}$
Moles of $SO_4$ = $\frac{1.92}{96} = 0.02 \text{ mol}$
Moles of $H_2O$ = $\frac{2.16}{18} = 0.12 \text{ mol}$

Determine Empirical Formula: Divide by the smallest number of moles (0.01):

K : Al : $SO_4$ : $H_2O$ = $1 : 1 : 2 : 12$
Empirical Formula = $KAl(SO_4)_2 \cdot 12H_2O$

Calculate Empirical Formula Mass: $\text{Mass} = 39 + 27 + 2(96) + 12(18) = 66 + 192 + 216 = 474 \text{ g mol}^{-1}$
Determine Molecular Formula:

$n = \frac{\text{Molar Mass}}{\text{Empirical Mass}} = \frac{948}{474} = 2$
Molecular Formula = $2 \times [KAl(SO_4)_2 \cdot 12H_2O] = K_2Al_2(SO_4)_4 \cdot 24H_2O$
This can be rewritten in the standard double salt notation as: $K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$ (Potash Alum).

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