States of Matter Quiz-5
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Q1. O2 is collected over water at 20℃. The pressure inside shown by the gas is 740 mm of Hg. What is the pressure due to O2 alone if vapour pressure of H2 O is 18 mm at 20℃ ?
Solution
(a) Pdry gas=Pwet gas-PH2 O
(a) Pdry gas=Pwet gas-PH2 O
Q2.Surface tension of water is 73 dyne cm-1 at 20℃. If surface area is increased by 0.10 m2, work done is
Solution
(b)
Work done = surface tension × increase in area
=73 dyne cm-1×0.10 m2
=73 dyne cm-1×0.10×104 cm2
=7.3×104 ergs
Q3. Which is lighter than dry air?
Solution
(a) Mol.wt. of moist air is lesser than dry air.
(a) Mol.wt. of moist air is lesser than dry air.
Q4. Which gas is most soluble in water?
Solution
(b) Due to H-bonding.
(b) Due to H-bonding.
Q5.Total energy of one mole of an ideal gas (monoatomic) at 27℃ is:
Solution
(b) KE=3/2 RT=3/2×2×300=900 cal
(b) KE=3/2 RT=3/2×2×300=900 cal
Q6. NH3 and HCl gas are introduced simultaneously from the two ends of a long tube. A white ring of NH4 Cl appears first
Solution
(a) White ring of NH4 Cl will appear nearer to the HCl end. The reason is that HCl (mol. wt. = 36.5) is heavier than NH3 (mol. wt. Hence, according to Graham’s law of diffusion, the rate of diffusion of NH3 will be higher than that of HCl.) r1/r2 =√(M2/M1 )
(a) White ring of NH4 Cl will appear nearer to the HCl end. The reason is that HCl (mol. wt. = 36.5) is heavier than NH3 (mol. wt. Hence, according to Graham’s law of diffusion, the rate of diffusion of NH3 will be higher than that of HCl.) r1/r2 =√(M2/M1 )
Q7.Non-reacting gases have a tendency to mix with each other. This property is known as:
Solution
(a)
Follow diffusion of gases.
Q8.Which gas cannot be kept in a glass bottle because it chemically reacts with glass?
Solution
(a) F2 is highly reactive gas.
(a) F2 is highly reactive gas.
Q9.The density of O2 is 16 at NTP. At what temperature its density will be 14? Consider that the pressure remains constant, at
Solution
(b)
d1 T1=d2 T2
When p remains constant
d1=16;
d2=14;T1=273 K,T2=?
d1 T1=d2 T2
16×273=14×T2
T2=312 K
T2=312-273=39℃
Q10. An open vessel at 27℃ is heated until 3/8th of the air in it has been expelled. Assuming that the volume remains constant, calculate the temperature at which the vessel was heated
Solution
(d)
According to Gay Lussac’s law
p1/T1 =p2/T2
If 3th/8 of the air is expelled out then remaining air =5/8
T2=((273+27)×8)/5
=2400/5=480 K
=480-273=207℃