A metal M forms the sulphate M2(S O4)3. A 0.596 gram sample of the sulphate reacts with excess BaCl2 to give 1.220 g BaSO4. What is the atomic weight of M (in g/mol )? (Molar mass of (BaSO)4 = 233 . 3 g / mol)

Question

A metal M forms the sulphate M2(S O4)3. A 0.596 gram sample of the sulphate reacts with excess BaCl2 to give 1.220 g BaSO4. What is the atomic weight of M (in g/mol )? (Molar mass of (BaSO)4 = 233 . 3 g / mol) (A) 26.9 (B) 69.7 (C) 55.8 (D) 23

Tardigrade Admin · Accepted Answer

The balanced equation for the given reaction is,

M_{2} (S O_{4})_{3} + 3 B a C l_{2} \to 2 MC l_{3} + 3 B a S O_{4}

Mole of

B a S O_{4} = \frac{1.22}{233.3} m o l

By stoichiometry of reaction:
Moles of

M_{2} (S O_{4})_{3} = \frac{1.22}{233.3} \times \frac{1}{3}

= 1.743 \times 1 0^{- 3} m o l

Mass of

M_{2} (S O_{4})_{3} = 0.596 g

∴ 1.743 \times 1 0^{- 3} (2 M + 96 \times 3) = 0.596 M = 26.9 g / m o l

Q. A metal $M$ forms the sulphate $M_{2} (S O_{4})_{3} .$ A $0.596$ gram sample of the sulphate reacts with excess $B a C l_{2}$ to give $1.220 g B a S O_{4} .$ What is the atomic weight of $M$ (in $g / m o l$ )? $(M o l a r ma sso f (B a SO)_{4} = 233.3 g / m o l)$

$26.9$

$69.7$

$55.8$

$23$

Q. A metal M forms the sulphate M2​(SO4​)3​. A 0.596 gram sample of the sulphate reacts with excess BaCl2​ to give 1.220gBaSO4​. What is the atomic weight of M (in g/mol )? (Molarmassof(BaSO)4​=233.3g/mol)

26.9

69.7

55.8

23

Q. A metal $M$ forms the sulphate $M_{2} (S O_{4})_{3} .$ A $0.596$ gram sample of the sulphate reacts with excess $B a C l_{2}$ to give $1.220 g B a S O_{4} .$ What is the atomic weight of $M$ (in $g / m o l$ )? $(M o l a r ma sso f (B a SO)_{4} = 233.3 g / m o l)$

$26.9$

$69.7$

$55.8$

$23$