Question

5. Assuming complete decomposition, the volume of Co, released at S.T.P. on heating 9.55 g Baco, will be (Atomic weight of Ba = 137) (1) 1.12 L (2) 0.84 L (3) 2.24 L no and 32 g HCI? (Molar mass (4) 4.06 L​

Answer 1

Question:-

Assuming complete decomposition, the volume of Co, released at S.T.P. on heating 9.55 g Baco, will be (Atomic weight of Ba = 137).

Given:-

• when heated BaCo = 9.55g.

• Atomic weight of Barium Carbonate = 137.

Solution:-

${\sf{{1 \: mole \: of \: BaCo_{3}}}}{\sf{{ \: gives = 22.4l \: Co_{2}}}}$

${\sf{{In \: question \: you \: have \: given \: 137 \: has \: _{}}}} \\ {\sf{{a \: weight \: of \: BaCo_{}}}}.$

In question mole of weight of Ba is given but not a Carbon and oxygen is given because they are common.

So,137 of Ba + 12 of Carbon + 16 × 3 of oxygen.

molecular weight of BaCo3 = molecular weight of Ba + molecular weight of C + 3 × molecular weight of O.

${\sf{{no. \: moles \: in \: Ba_{2}}}}{\sf{{Co_{3}}}}$

$= \frac{9.55}{197} \: \\ \\ here \: 197 \: is \: molecule \: weight \: of \: \\ {\sf{{ BaCo_{3}}}}.$

$\frac{9.55}{197} \: moles \: will \: be \\ \\ = \frac{9.55}{197} \times 22.4l$

Answer:- = 1.085l