An electric discharge is passed through a mixture containing 50 cc of $O_{2}$ and 50 cc of $H_{2}$ . The volume of the gases formed (i) at room temperature, (ii) at 110C will be:

(i) 25 cc (ii) 50 cc

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

(i) 50 cc (ii) 75 cc

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

(i) 25 cc (ii) 75 cc

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

(i) 75 cc (ii) 75 cc

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

Open in App

Solution

The correct option is B (i) 25 cc (ii) 75 cc
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O$
50 cc H2 will combine with 25 cc $O_{2}$ to form 50 cc $H_{2} O$
$∴ O_{2}$ left = 25 cc
At room temperature, $H_{2} O$ will be in liquid state but at $110^{o} C$ , it will be gaseous. Thus, volume of gases at 25C and $110^{o} C$ will be 25 cc and 75 cc respectively.

Suggest Corrections

Similar questions

Q. An electric discharge is passed through a mixture contaning

50

cc of

O_{2}

and

50

cc of

H_{2}

. The volume of the gases formed (i) at room temperature and (ii) at

110^{0} C

will be, respectively:

An electric discharge is passed through a mixture containing 50 cc of $O_{2}$ and 50 cc of $H_{2}$ . The volume of the gas formed (i) at room temperature (ii) at $110^{\circ} C$ will be:

(All measurements are made at same pressure and temperature)

Q. An electric discharge is passed through a mixture containing 50 c.c. of

O_{2}

and 50 c.c. of

H_{2}

. The volume of the gases formed (i) at room temperature and (ii) at 110⁰C will be

An electric discharge is passes through a mixture containing 50 c.c. of $H_{2}$ and 50 c.c. of $O_{2}$ . The volume of the gases formed (i) at room temperature and (ii) at $110^{\circ} C$ will be:

Q. At electric discharge is passed through a mixture containing 50 cc of

O_{2}

and 50 cc of

H_{2}

. The volume of the gas removed at room temperature will be, respectively:

Join BYJU'S Learning Program

An electric discharge is passed through a mixture containing 50 cc of O2 and 50 cc of H2. The volume of the gases formed (i) at room temperature, (ii) at 110C will be:

The correct option is B (i) 25 cc (ii) 75 cc2H2(g)+O2(g)→2H2O 50 cc H2 will combine with 25 cc O2 to form 50 cc H2O∴O2 left = 25 cc At room temperature, H2O will be in liquid state but at 110oC, it will be gaseous. Thus, volume of gases at 25C and 110oCwill be 25 cc and 75 cc respectively.

An electric discharge is passed through a mixture containing 50 cc of $O_{2}$ and 50 cc of $H_{2}$ . The volume of the gases formed (i) at room temperature, (ii) at 110C will be: