Question bank/JEE Main/Chemistry

JEE MainChemistryToughPhysical Chemistry

The minimum volume of the water needed to dissolve 1 g of BaSO4
(Ksp = 10-10) is about: Assume no reaction of cation/anion.
[Mol. mass (BaSO4) = 233 u]
1g BaSO4 (Ksp =10-10) dks ?kksyus ds fy, ty
yxHkx........

Show answer and solution
Solubility of BaSO_4 =
K5p=10−10=1×10−5M\sqrt{K_{5p}} = \sqrt{10^{- 10}} = 1 \times 10^{- 5}M
∴\therefore wt. of BaSO4 = 1 × 10-5 × 233 = 233 × 10-5 g/litre.
∴\therefore Volume of water needed to dissolve 1 g BaSO4 is equal to
1233×10−5≈\frac{1}{233 \times 10^{- 5}} \approx 430 litres
BaSO_4 =
K5p=10−10=1×10−5M\sqrt{K_{5p}} = \sqrt{10^{- 10}} = 1 \times 10^{- 5}M
∴\therefore BaSO4 = 1 × 10-5 × 233 = 233 × 10-5 xzke/yhVj.
∴\therefore 1gBaSO4 dks ?kksyus ds fy, ty

Source: Grass-G / CH-26 Chemistry Paper 1 15 Jan.docx

More JEE Main questionsChapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubtView plans