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New answer posted
3 months agoContributor-Level 10
∴ % of S in the compound
= (32/233) * (mass of BaSO? / mass of compound) * 100 = (32 * 0.35 * 100) / (233 * 0.25) = 19.227 ≈ 19.23
New answer posted
3 months agoContributor-Level 10
? Solvent is H? O, which is in excess
So using m ( molality ) = (x? *1000)/ (x? * (M? )? )
? x? = 0.74 (Mol? = 18 g)
x? = 1 – 0.74 = 0.26 ∴ m = (0.26 * 1000)/ (0.74 * 18) = 19.5
New answer posted
3 months agoNew answer posted
3 months agoContributor-Level 10
ppm of O? = (wt. of O? ) / (wt. of H? O) * 10?
= (10.3 mg) / (1.03 * 10? mg) * 10?
= 10ppm
New answer posted
3 months agoContributor-Level 10
i = total no. of particles after dissociation/association / total no. of particle before dissociation/association
HA? H? + A? (Dissociation)
0.5a
2HA → (HA)? (Association)
0.5a/2
i = (a + 0.25a)/a = 125 x 10? ²
Ans. = 125
New question posted
3 months agoNew answer posted
3 months agoContributor-Level 10
Molality = moles of solute / mass of solvent (kgs) ⇒ 1 = 0.5 / W (kgs)
Wsolvent (kgs) = 0.5 = 500 g
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