A 2.24 L cylinder of oxygen at N.T.P. is found to develop a leakage. When the leakage was plugged, the pressure dropped to 570 mm of Hg. The number of moles of gas that escaped will be

Question

A 2.24 L cylinder of oxygen at N.T.P. is found to develop a leakage. When the leakage was plugged, the pressure dropped to 570 mm of Hg. The number of moles of gas that escaped will be (A) 0.025 (B) 0.050 (C) 0.075 (D) 0.09

Tardigrade Admin · Accepted Answer

No. of moles originally present =(2.24/22.4)=0.1 After leakage, P V=n R T or n=(P V/R T)=((570 / 760)(2.24)/0.0821 × 273)=0.075 Hence, no. of moles leaked out =0.1-0.075=0.025 mol

Q. A $2.24 L$ cylinder of oxygen at $N . T . P$ . is found to develop a leakage. When the leakage was plugged, the pressure dropped to $570 mm$ of $H g$ . The number of moles of gas that escaped will be

$0.025$

$0.050$

$0.075$

$0.09$

No. of moles originally present $= \frac{2.24}{22.4} = 0.1$

After leakage, $P V = n RT$

or $n = \frac{P V}{RT} = \frac{( 570/760 ) ( 2.24 )}{0.0821 \times 273} = 0.075$

Hence, no. of moles leaked out $= 0.1 - 0.075 = 0.025 m o l$

Q. A 2.24L cylinder of oxygen at N.T.P. is found to develop a leakage. When the leakage was plugged, the pressure dropped to 570mm of Hg. The number of moles of gas that escaped will be

0.025

0.050

0.075

0.09