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`        One gram of an alloy of aluminium and magnesium when treated with excess of dil. HCI forms magnesium chloride, aluminium chloride and hydrogen. The evolved hydrogen, collected over mercury at 0oC has a volume of 1.20 litres at 0.92 atm. Pressure. Calculate the composition of the alloy[H = 1, Mg = 24, Al = 27]`
5 years ago

Jitender Pal
365 Points
```
(i) Find volume of H2 at N.T.P.

(ii) Total amount of H2 liberated = H2 liberated by Mg & HCI + H2 liberated by Al & HCI.

Conversion of volume of H2 to N.T.P

Given condition N.T.P condition

P1= 0.92 atm. P2 = 1 atm.

V1 = 1.20 litres V2= ?

T1 = 0 + 273 = 273 K T2 = 273 K

Applying ideal gas equation, P1V1/T1 = P2V2/T2

0.92 *1.20/273 = 1 *V2/273 ,V2 0.92 * 1.20 *273/273 *1 litres = 1.104

Litres = 1104 ml

The relevant chemical equation are

(i) 2 Al + 6HCI → 2AICI3 + 3H2

2 * 27  3 *22400

= 54 g = 67200 ml at NTP

(ii) Mg + 2HCI → MgCI2 + H2

24 g 22400 ml at NTP

Wt. of alloy = 1g

Let the wt. of aluminium in alloy = x g

∴ Wt. of magnesium in alloy = (1 -x) g

According to equation (i)

54 g Al = 67200 ml of H2 at N.T.P

∴ x g of Al = 67200/54 * x = 1244.4 x ml of H2 at N.T.P

Similarly, from equation (ii)

24 g of Mg = 222400 ml of H2 at N.T.P

(1 - x) g of Mg = 222400/24 * (1 - x) = 933.3 (1 - x) ml of H2

Hence total vol. of H2 collected at N.T.P

= 12244.4 x + 933.3 (1 - x)ml

But total vol. of H2 as calculated above = 1104 ml

∴ 1244.4 x + 933.3 (1 - x) = 1104 ml

1244.4 x – 933.3 x = 1104 – 933.3

311.1 x = 170.7, x = 0.5487

Hence 1 g of alloy contains Al = 0.5487 g

∴ percentage of Al in alloy = 0.5487 *100/1 = 54.87%

% of Mg in alloy = 100 – 54.87 = 45.13%

```
5 years ago
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