2.9 g of a gas at `95^@` occupied the same volume as 0.184 g of dihydrogen at `17^@C`,at the same pressure, what is the molar mass of the gas?

Explanation

Given unknown gas and dihydrogen
For unknown gas :
` V_1 = V`
` n_1 = ( 2.9 )/(m )`
` T_1 = 95^@ C=368 K`
` P_1 V_1 = n_1 RT_1`
` P_1 = (n_1 RT_1)/(V_1)`
` = ( 2.9 )/( m ) xx ( R xx 368 )/ (V ) `
for dihydrogen
` V_2 =V_0`
` n_2 = ( 0.184)/(2)`
`T_2 = 17^@C = 290 K`
` P_2 V_2 = n_2 RT_2`
` P_2 = (n_2 RT_2)/( V_2)`
` = ( 0.184 )/(2) xx (R xx 290 )/(V )`
Given pressures are same
` therefore (2.9 )/( m ) xx (R xx 368 ) /(V ) = (0.184)/(2) xx (R xx 290)/(V)`
` therefore ( 2.9 xx 368 )/( 0.184 xx 145 ) = 40 gms // ml`