Question

Which has maximum volume at S.T.P.?
(a) 1.5 * 1023 molecules of CO,
(b) 1 g H2
(c)4 902
(d) 16 g So,​

Answer 1

Which has maximum volume at S.T.P.?

(a) 1.5 * 1023 molecules of CO,

(b) 1 g H2✔

(c)4 902

(d) 16 g So,

Answer 2

Answer:

(b)Out of all of these,1gm of $H_{2}$ occupies maximum volume at S.T.P

Explanation:

The correct question is-

Which has maximum volume at S.T.P.?

(a) $1.5\times10^{23}$ molecules of CO

(b) 1 g H2

(c)4g $O_{2}$

(d) 16g SO

We know that one mole of any gas occupies a volume of 22.4L at Standard temperature and pressure.Hence,we calculate the volumes of the given gases as shown below-

(a) $1.5\times10^{23}$ molecules of CO

Number of moles of given gas is the ratio of given molecules to the Avagadro number.Therefore,

$n=\frac{m}{N_{A}} =\frac{1.5\times10^{23}}{6.023\times10^{23}} =0.249mol$

As 1 mole occupies 22.4L,the volume occupied by 0.249mol is

$V=0.249\times22.4=5.57L$

Hence the volume occupied is 5.57L

(b) 1 g H2

Number of moles of hydrogen is the ratio of its given mass and molar mass.Therefore,

$n=\frac{1}{2} =0.5mol$

As 1 mole occupies 22.4L,the volume occupied by 0.5mol is

$V=0.5\times22.4=11.2L$

(c)4g $O_{2}$

Number of moles of Oxygen is the ratio of its given mass and molar mass.Therefore,

$n=\frac{4}{32} =0.125mol$

As 1 mole occupies 22.4L,the volume occupied by 0.125mol is

$V=0.125\times22.4=2.8L$

(d) 16g SO

Number of moles of sulphur oxide is the ratio of its given mass and molar mass.Therefore,

$n=\frac{16}{48} =0.33mol$

As 1 mole occupies 22.4L,the volume occupied by 0.33mol is

$V=0.33\times22.4=7.39L$

Therefore,

Out of all of these,1gm of $H_{2}$ occupies maximum volume at S.T.P

#SPJ2