The volume of water needed to dissolve 1 mg of P bSO 4 K sp -1-44 - 10-8 M 2 at 25- C is approximately-molar mass of Pb -207 g - mol A- 10 mLB- 27 mLC- 43 mLD- 80 mL

The correct option is B 27 mLSolubility of PbSO_4 = nbsp;√(K_sp) = √(1.44 × 10^ 8) = 1.2 × 10^ 4 M Solubility of PbSO_4 = 1.2 × 10^ 4 nbsp;× 303 × 10^3 = ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Solubility Product

The volume of...

Question

The volume of water needed to dissolve $1 mg$ of $P b S O_{4} (K_{s p} = 1.44 \times 10^{- 8} M^{2})$ at $25^{\circ} C$ is approximately:
(molar mass of $P b = 207 g/mol$ )

10 mL

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

27 mL

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

43 mL

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

80 mL

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

Open in App

Solution

The correct option is B $27 mL$
$Solubility of P b S O_{4} = \sqrt{K_{s p}} = \sqrt{1.44 \times 10^{- 8}} = 1.2 \times 10^{- 4} M Solubility of P b S O_{4} = 1.2 \times 10^{- 4} \times 303 \times 10^{3} = 36.36 {mg L}^{- 1} ∴ Volume of water needed to dissolve 1 mg of P b S O_{4} = \frac{1000}{36.36} = 27.5 mL$

Suggest Corrections

Similar questions

Q. The volume of water needed to dissolve

1 mg

P b S O_{4} (K_{s p} = 1.44 \times 10^{- 8} M^{2})

25^{\circ} C

is approximately:
(molar mass of

P b = 207 g/mol

)

Q. The volume of the water needed to dissolve

1

g of

B a {S O}_{4}

(

K_{s p} = 1.1 \times 10^{- 10}

) at

25^{o} C

is:

Q. The minimum volume of water required to dissolve

0.1

g lead(II) chloride to get a saturated solution (

K_{s p}

P b C l_{2} = 3.2 \times 10^{- 8}

; atomic mass of Pb

= 207 u

) is:

Q. At

25^{o}

C, required volume of water, to dissolve

1

B a S O_{4}

(K_{s p} = 1.1 \times 10^{- 10})

will be:

[Molecular weight of

B a S O_{4} = 233

]

Q. If the degree of dissociation of water at

70^{o} C

1.28 \times 10^{- 8}

. Then what is the ionic product of water at this temperature?

Join BYJU'S Learning Program

The volume of water needed to dissolve 1 mg of PbSO4(Ksp=1.44×10−8M2) at 25 ∘C is approximately: (molar mass of Pb=207 g/mol)

The correct option is B 27 mLSolubility of PbSO4=√Ksp=√1.44×10−8=1.2×10−4 MSolubility of PbSO4=1.2×10−4×303×103=36.36 mg L−1∴Volume of water needed to dissolve 1 mg of PbSO4=100036.36=27.5 mL

The volume of water needed to dissolve $1 mg$ of $P b S O_{4} (K_{s p} = 1.44 \times 10^{- 8} M^{2})$ at $25^{\circ} C$ is approximately:
(molar mass of $P b = 207 g/mol$ )