0-303 g of the organic compound in Duma-s method on combustion gave 32-27 ml of moist nitrogen at 27-C and at P mm of Hg pressure- The vapour pressure at 27-C is 21 mm of Hg- Find the value of P-

The correct option is B 746 mm of HgAccording to the standard equation, P_0V_0T_0 = P_1V_1/T_1 Substituting values in the above expression, we get ...

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

Byju's Answer

Standard XII

Chemistry

Concentration Terms - An Introduction (w/w etc)

0.303 g of th...

Question

$0.303$ g of the organic compound in Duma's method on combustion gave $32.27$ ml of moist nitrogen at $27^{\circ}$ C and at P mm of Hg pressure. The vapour pressure at $27^{\circ}$ C is $21$ mm of Hg. Find the value of P.

725

mm of Hg

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

746

mm of Hg

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

710

mm of Hg

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

760

mm of Hg

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

Open in App

Solution

The correct option is B $746$ mm of Hg
According to the standard equation, $\frac{P_{0} V_{0}}{T_{0}} = \frac{P_{1} V_{1}}{T_{1}}$

Substituting values in the above expression, we get

$\frac{760 \times 26.92}{273} = \frac{P_{1} \times 32.27}{298}$

Hence, $P_{1}$ is $746$ mm of Hg.

Suggest Corrections

Similar questions

Q. In Duma's method of estimation of nitrogen, 0.1840 g of an organic compound gave 30 mL of nitrogen collected at 287 K and 758 mm of Hg pressure. The percentage composition of nitrogen in the compound is ___. (Round to the nearest integer). [Given: Aqueous tension at 287 K = 14 mm of Hg]

0.3 g

of an organic compound gives

50 mL

N_{2}

gas at

27^{\circ} C

and

715 mm

of Hg pressure. Calculate the percentage of

N_{2}

, given that the aqueous tension at

27^{\circ} C

15 mm

H g

Q. At constant temperature

250 m L

of argon at a

760 m m H g

pressure and

60 m L

of nitrogen at

500 m m

pressure are put together in one litre flask. The final pressure is _________

m m H g

Q. The vapour pressure of mercury is

0.002 m m H g

27^{o} C . H g (l) ⇋ H g (g),

the value of equilibrium constant

K_{c}

is :

Q. Lowering of vapour pressure due to a solute in 1 molal aqueous solution at

100^{\circ} C

Join BYJU'S Learning Program

0.303 g of the organic compound in Duma's method on combustion gave 32.27 ml of moist nitrogen at 27∘C and at P mm of Hg pressure. The vapour pressure at 27∘C is 21 mm of Hg. Find the value of P.

The correct option is B 746 mm of HgAccording to the standard equation, P0V0T0=P1V1T1Substituting values in the above expression, we get760×26.92273=P1×32.27298Hence, P1 is 746 mm of Hg.

$0.303$ g of the organic compound in Duma's method on combustion gave $32.27$ ml of moist nitrogen at $27^{\circ}$ C and at P mm of Hg pressure. The vapour pressure at $27^{\circ}$ C is $21$ mm of Hg. Find the value of P.

The correct option is B $746$ mm of Hg
According to the standard equation, $\frac{P_{0} V_{0}}{T_{0}} = \frac{P_{1} V_{1}}{T_{1}}$

Substituting values in the above expression, we get

$\frac{760 \times 26.92}{273} = \frac{P_{1} \times 32.27}{298}$

Hence, $P_{1}$ is $746$ mm of Hg.