Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 2.9 g of a gas at  occupied the same volume as 0.184 g of dihydr

2.9 g of a gas at \mathrm{95^{\circ}C} occupied the same volume as 0.184 g of dihydrogen at \mathrm{17^{\circ}C} , at the same pressure . which is the molar mass of the gas _________ 

Option: 1

30 g/mol 


Option: 2

60 g/mol


Option: 3

20 g/mol


Option: 4

40 g/mol


Answers (1)

best_answer

Let the molar mass of the unknown gas =M g/mol

Number of molar for 2.9 g of the gas \mathrm{=\frac{2.9}{M}mol}

Number of moles for 0.184 g of \mathrm{H_{2 }} gas \mathrm{=\frac{0.184}{2}=0.092 \mathrm{~mol} .}

As both the gases have same volume at the same pressure and specified temperature , they will have the same value of PV .

To calculate PV, we apply the equation,

\mathrm{P V=n R T}\mathrm{P V=\frac{2.9}{M} R(213+95) [faro\; unknown \; g \; as ]}\mathrm{P V=\frac{2.9}{M} R(213+95) [faro\; unknown \; g as ]}

\mathrm{P V=0.092 \mathrm{R}(273+17)\left[\text { for } \mathrm{H}_2 \text { gas }\right]}

Therefore, \mathrm{\begin{aligned} & \frac{2.9}{M} R \times 368=0.092 R \times 290 \\ & M=40 \mathrm{~g} / \mathrm{mol} \end{aligned}}

Posted by

Gunjita

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 session 1 exam in the last week of January; registration details
anam-khan

Over 656 UPSC, JEE, NEET, other coaching aspirants refunded more than Rs 1 crore: Govt
anam-khan

NTA Exam Calendar 2025: JEE Main, NEET, CUET, UGC NET dates soon; websites to check

Latest Question