54168.The number of water molecules is maximum in
1.8 gram of water
18 gram of water
18 moles of water
18 molecules of water.
Explanation:
As learnt in
>Mole Concept -One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope.
- wherein
one mole = 6.0221367 X 10^23
Let us check the number of molecules in each option.
18 molecules of water have 18 molecules
1.8 gram of water has $\dfrac{1.8}{18}\times N_{A}$ molecules
18 gram of water has $\dfrac{18}{18} N_{A}$ molecules
18 moles of water has 18 NA molecules
As learnt in
>Mole Concept -One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope.
- wherein
one mole = 6.0221367 X 10^23
Let us check the number of molecules in each option.
18 molecules of water have 18 molecules
1.8 gram of water has $\dfrac{1.8}{18}\times N_{A}$ molecules
18 gram of water has $\dfrac{18}{18} N_{A}$ molecules
18 moles of water has 18 NA molecules
54169.Suppose the elements X and Y combine to form two compounds XY2 and X3Y2 . When 0.1 mole of XY2 weighs 10 g and 0.05 mole of X3Y2 weighs 9 g, the atomic weights of X and 7 are
40,30
60,40
20,30
30,20
Explanation:
Molar Mass - The mass of one mole of a substance in grams is called its molar mass.
wherein-Molar mass of water = 18 g mol-1
Let the atomic weight of X = x and atomic weight of Y = y
then , x+2y = $\dfrac{10}{0.1}$
and 3x+2y = $\dfrac{9}{0.05}$ = 180
as saluiy , we get
x=40,y=30
Molar Mass - The mass of one mole of a substance in grams is called its molar mass.
wherein-Molar mass of water = 18 g mol-1
Let the atomic weight of X = x and atomic weight of Y = y
then , x+2y = $\dfrac{10}{0.1}$
and 3x+2y = $\dfrac{9}{0.05}$ = 180
as saluiy , we get
x=40,y=30
54170.Equal masses of H2 , O2 and methane have been taken in a container of volume V at
temperature 27 °C in identical conditions. The ratio of the volumes of gases H2 : O2 : methane would be
temperature 27 °C in identical conditions. The ratio of the volumes of gases H2 : O2 : methane would be
8 : 16 : 1
16 : 8 : 1
16 : 1 : 2
8 : 1 : 2
Explanation:
As learnt in Partial Pressure of a gas in gaseous mixture -
Partial Pressure = (Total Pressure of Mixture) $ \times $ (mole fraction of gas)
Suppose mass is m
moles of $ H_{2} $ = m/2
moles of $ O_{2} $= m/32
moles of methane = m/16
Volume ratio = $ \dfrac{m}{2}=\dfrac{m}{32}=\dfrac{m}{16} $
= 16:1:2
As learnt in Partial Pressure of a gas in gaseous mixture -
Partial Pressure = (Total Pressure of Mixture) $ \times $ (mole fraction of gas)
Suppose mass is m
moles of $ H_{2} $ = m/2
moles of $ O_{2} $= m/32
moles of methane = m/16
Volume ratio = $ \dfrac{m}{2}=\dfrac{m}{32}=\dfrac{m}{16} $
= 16:1:2
54171.If Avogadro number N , is changed from 6.022 x 1023 mol-1 to 6.022 x 10^20 mol , this would change
the mass of one mole of carbon
the ratio of chemical species to each other in a balanced equation
the ratio of elements to each other in a compound
the definition of mass in units of grams.
Explanation:
As we learnt in Mole Concept
One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope.
wherein
one mole = 6.0221367 X 10^23
So if now, I mole = 6.022X10^20
then 6.022X10^20 atoms of carbon mass = $\dfrac{12g}{1000}$
1 mol of carbon atom mass = 0.012g
As we learnt in Mole Concept
One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope.
wherein
one mole = 6.0221367 X 10^23
So if now, I mole = 6.022X10^20
then 6.022X10^20 atoms of carbon mass = $\dfrac{12g}{1000}$
1 mol of carbon atom mass = 0.012g
54172.A mixture of gases contains H2 and O2 gases in the ratio of 1 : 4 (w/w). What is the molar ratio of the two gases in the mixture?
16:1
2 : 1
1 : 4
4 : 1
Explanation:
Partial Pressure of a gas in gaseous mixture -
Partial Pressure = (Total Pressure of Mixture) $ \times $ (mole fraction of gas)
suppose mass of $ H_{2} $ = x
then, mass of $ O_{2} $ = 4x
Moles of $ H_{2}=\dfrac{x}{2} $
Moles of $ O_{2}=\dfrac{4x}{32}=\dfrac{x}{8}$
molar ratio = $ \dfrac{\dfrac{x}{2}}{\dfrac{x}{8}} $ = 4:1
Partial Pressure of a gas in gaseous mixture -
Partial Pressure = (Total Pressure of Mixture) $ \times $ (mole fraction of gas)
suppose mass of $ H_{2} $ = x
then, mass of $ O_{2} $ = 4x
Moles of $ H_{2}=\dfrac{x}{2} $
Moles of $ O_{2}=\dfrac{4x}{32}=\dfrac{x}{8}$
molar ratio = $ \dfrac{\dfrac{x}{2}}{\dfrac{x}{8}} $ = 4:1
54173.1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel. Which reactant is left in
excess and how much? (At. wt. Mg = 24, O=16)
excess and how much? (At. wt. Mg = 24, O=16)
Mg, 0.16 g
0 , 0.16 g
Mg, 0.44 g
0 0.28 g
Explanation:
The reaction is as follows
$ M_{g}+\dfrac{1}{2}O_{2}\rightarrow M_{g}O $
modes of $ M_{g}=\dfrac{1}{24}=0.04167 $
modes of $ O_{2}=\dfrac{0.56}{32}=0.0175 $
According to reaction stoichiometry $ \dfrac{1}{24 } $ modes of $ M_{g} $ require $ \dfrac{1}{48} $
modes of $ O_{2} $ or 0.02083 modes of $ O_{2} $
clearly $ O_{2} $ is the limiting reagent modes of $ M_{g} $ required by 0.01175 modes of $ O_{2} $
=0.0175 $ \times $ 2=0.035ml
$ \therefore $ Excess $ M_{g} $ =0.04167-0.035
=0.00667ml
=0.16 g
54174.In an experiment it showed that 10 mL of 0.05 M solution of chloride required 10 mL of 0.1
M solution of AgNO3 , which of the following will be the formula of the chloride (X stands for
the symbol of the element other than chlorine)
M solution of AgNO3 , which of the following will be the formula of the chloride (X stands for
the symbol of the element other than chlorine)
X2Cl2
XCl2
XCl2
X2Cl
Explanation:
Stoichiometry -
Stoichiometry deals with measurements of reactants and products in a chemical reaction.
- wherein
aA (g) + bB (g) → cC (g) + dD (g)
Here, ‘a’ moles of A(g) reacts with ‘b’ moles of B(g) to give ‘c’ mole of C(g) and ‘d’ moles of D(g)
3=10^(-3) mol
No. of moles the chloride= 0.5 x 10^(-3) mol
Suppose the formula for the chloride is XCln then moles of chloride ion=n x 0.5 x 10^(-3)
Reaction goes as follows:
>$ Ag^{+} + Cl^{-} \rightarrow AgCl $
Then, going by stoichiometry we get
n x 0.5 x 10^-3= 10^-3
$ \Rightarrow $ n=2
Therefore, formula is XCl2
Stoichiometry -
Stoichiometry deals with measurements of reactants and products in a chemical reaction.
- wherein
aA (g) + bB (g) → cC (g) + dD (g)
Here, ‘a’ moles of A(g) reacts with ‘b’ moles of B(g) to give ‘c’ mole of C(g) and ‘d’ moles of D(g)
No. of moles the chloride= 0.5 x 10^(-3) mol
Suppose the formula for the chloride is XCln then moles of chloride ion=n x 0.5 x 10^(-3)
Reaction goes as follows:
>$ Ag^{+} + Cl^{-} \rightarrow AgCl $
Then, going by stoichiometry we get
n x 0.5 x 10^-3= 10^-3
$ \Rightarrow $ n=2
Therefore, formula is XCl2
54175.6.02 x 10^20 molecules of urea are present in 100 mL of its solution. The concentration of
solution is
solution is
0.001 M
0.1 M
0.02 M
0.01 M
Explanation:
As learnt in
Molarity -
Molarity (M) = (Number of moles of solute)/(volume of solution in litres)
- wherein
It is defined as the number of moles of the solute in 1 litre of the solution.
No. of molecules (given) = $ 6.02 \times 10^{20} $
No. of moles = $ \dfrac{6.02 \times 10^{20}}{6.02\times 10^{23}}= 10^{-3} $ mol
Volume of solution=100 ml= 0.1 L
Therefore, Molarity=$ \dfrac{no.of moles}{volume} $
$ \dfrac{10^{-3}}{0.1} $ = 0.01 M
As learnt in
Molarity -
Molarity (M) = (Number of moles of solute)/(volume of solution in litres)
- wherein
It is defined as the number of moles of the solute in 1 litre of the solution.
No. of molecules (given) = $ 6.02 \times 10^{20} $
No. of moles = $ \dfrac{6.02 \times 10^{20}}{6.02\times 10^{23}}= 10^{-3} $ mol
Volume of solution=100 ml= 0.1 L
Therefore, Molarity=$ \dfrac{no.of moles}{volume} $
$ \dfrac{10^{-3}}{0.1} $ = 0.01 M
54176.When 22.4 litres of H2(g) is mixed with 11.2 litres of Cl2(g) each at S.T.P, the moles of
HCl(g) formed is equal to
HCl(g) formed is equal to
1 mol of HCl
2 mol of HCl
0.5 mol of HCl
1.5 mol of HCl
Explanation:
H2 + CL2-> 2HCL 22.4 L + 22.4 L -> 2×22.4
22.4 Litres hydrogen reacts with 22.4 litre of chlorine to give 44.8 L of HCL
the given volume of chlorine is 11.2 L
here chlorine is the liming reagents as its . first consumed
22.4 Litre of chlorine gives 73 litre of HCL
so , 11.2 l gives 44.8/22.4 × 11.2 = 22.4 of HCL is formed
that means no. of mole = 1
because 1 mole of any gases occupies 22.4 L at STP
the answer is A . 1 mole of HCL (g)
H2 + CL2-> 2HCL 22.4 L + 22.4 L -> 2×22.4
22.4 Litres hydrogen reacts with 22.4 litre of chlorine to give 44.8 L of HCL
the given volume of chlorine is 11.2 L
here chlorine is the liming reagents as its . first consumed
22.4 Litre of chlorine gives 73 litre of HCL
so , 11.2 l gives 44.8/22.4 × 11.2 = 22.4 of HCL is formed
that means no. of mole = 1
because 1 mole of any gases occupies 22.4 L at STP
the answer is A . 1 mole of HCL (g)
54177.What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO3 is
mixed with 50 ml of 5.8% NaCl solution? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5)
mixed with 50 ml of 5.8% NaCl solution? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5)
3.5 g
7 g
14 g
28 g
Explanation:
As we discussed in Mass percent of solution Mass percent = ((mass of solute)/(mass of solution) ) X 100wherein
it is mass of solute present in 100 gram solution.16.9% solution of AgNO3 means 16.9g AgNO3 in 100ml solution.
50ml of solution contains $\dfrac{16.9}{2}$=8.45g
Similarly, 100ml of other solution contains 5.8g NaCl
50ml of other solution contains 2.9g NaCl
Mol of AgNO3 = $\dfrac{8.45}{169.8}$=0.049
Mol of NaCl = $\dfrac{2.9}{58.5}$=0.049
AgNO3 + NaCl $\rightarrow$ Ag03Cl + Na0NO3
Mass of AgCl = 0.049 x Molar mass of AgCl
= 0.049 x 143.5 = 7g
As we discussed in Mass percent of solution Mass percent = ((mass of solute)/(mass of solution) ) X 100wherein
it is mass of solute present in 100 gram solution.16.9% solution of AgNO3 means 16.9g AgNO3 in 100ml solution.
50ml of solution contains $\dfrac{16.9}{2}$=8.45g
Similarly, 100ml of other solution contains 5.8g NaCl
50ml of other solution contains 2.9g NaCl
Mol of AgNO3 = $\dfrac{8.45}{169.8}$=0.049
Mol of NaCl = $\dfrac{2.9}{58.5}$=0.049
AgNO3 + NaCl $\rightarrow$ Ag03Cl + Na0NO3
0.049 0.049 0.049 0.049 0 0
Mass of AgCl = 0.049 x Molar mass of AgCl
= 0.049 x 143.5 = 7g
54178.The resistance of material depends on :
temperature
length of conductor
area of cross-section
All the above
54179.The physical quantity which does not have the same dimensions as the other three is
spring constant
surface tension
surface energy
acceleration due to gravity
54180.Newton - second is the unit of
Velocity
Anguler momentum
Momentum
Energy
Explanation:
Impulse = change in momentum = F × t So the unit of momentuim will be equal to Newton− sec)
Impulse = change in momentum = F × t So the unit of momentuim will be equal to Newton− sec)
54181.Electric current originates from which part of an atom?
nucleus
entire atom acting as a unit
positively charged protons
negatively charged electrons
54182.Which of the following is not the unit of energy ?
calorie
joule
electron volt
watt
Explanation:
watt is a unit of power
watt is a unit of power
54183.When a person combs her hair, static electricity is sometimes generated by what process?
Friction between the comb and hair transfers electrons.
Induction between the comb and hair.
Deduction between the comb and hair.
Contact between the comb and hair results in a charge.
54184.The numerical value of a given quantity is
independent of unit
directly proportional to unit
inversely proportional to unit
directly proportional to the square root of the unit
Explanation:
The numerical value of a given quantity is inversely proportional to unit.
The numerical value of a given quantity is inversely proportional to unit.
54185.Unit of reduction factor is
ampere
ohm
tesla
weber
Explanation:
Reduction factor of say tangent galvanometer is actually numerically equal to the current in ampere needed to produce a deflection of 45°, when plane of coil lies in magnetic meridian.
=> $1 = K \phi$ where,
K is a constant called reduction factor. Its unit is same as of current that is ampere.
Reduction factor of say tangent galvanometer is actually numerically equal to the current in ampere needed to produce a deflection of 45°, when plane of coil lies in magnetic meridian.
=> $1 = K \phi$ where,
K is a constant called reduction factor. Its unit is same as of current that is ampere.
54187.If h is Planck’s constant and X is wavelength, h/X has dimensions of
momentum
energy
mass
velocity
Explanation:
We know that,
$\lambda=\frac{h}{mv}$
$\therefore \frac{h}{\lambda}= mv$
Hence, $\frac{h}{\lambda}$ has dimensions of momentum.
We know that,
$\lambda=\frac{h}{mv}$
$\therefore \frac{h}{\lambda}= mv$
Hence, $\frac{h}{\lambda}$ has dimensions of momentum.