54208.Which has the maximum number of molecules among the following?
44g CO2
48g O3
8g HO2
64g SO2
Explanation:
The number of moles can be calculated from the ratio of the mass to the molar mass.
Maximum number of moles will corresponds to maximum number of molecules.
The number of moles can be calculated using the formula,
No.of moles = Weight of the substance/Molecular weight of the substance
Moles of CO2 = 44/44 = 1 mol.
Moles of O3 = 48/48 = 1 mol
Moles of H2 = 8/2 = 4 mol.
Moles of SO2 = 64/64 = 1 mol.
4 moles of H2, i.e. 4×6.023×1023 molecules.
Hence, 8g H2 has the maximum number of molecules.
The number of moles can be calculated from the ratio of the mass to the molar mass.
Maximum number of moles will corresponds to maximum number of molecules.
The number of moles can be calculated using the formula,
No.of moles = Weight of the substance/Molecular weight of the substance
Moles of CO2 = 44/44 = 1 mol.
Moles of O3 = 48/48 = 1 mol
Moles of H2 = 8/2 = 4 mol.
Moles of SO2 = 64/64 = 1 mol.
4 moles of H2, i.e. 4×6.023×1023 molecules.
Hence, 8g H2 has the maximum number of molecules.
54209.The number of atoms in 0.1 mol of a triatomic gas is (N = 6.02 x 10^23 mol )
6.026 x 10^22
1.806 x 10^23
3.600 x 10^23
1.800 x 10^22
Explanation:
Number of moles in 0.1 mole of triatomic gas = 0.1 x 3 x 6.02 x 10^23
=1.806 x 10^23
Number of moles in 0.1 mole of triatomic gas = 0.1 x 3 x 6.02 x 10^23
=1.806 x 10^23
54210.25.3 g of sodium carbonate, Na2CO3 is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion,
Na+ and carbonate ions, CO32- are respectively (Molar mass of Na2CO3 = 106g mol-1
Na+ and carbonate ions, CO32- are respectively (Molar mass of Na2CO3 = 106g mol-1
0.955 M and 1.910 M
1.910 M and 0.955 M
1.90 M and 1.910 M
0.477 M and 0.477 M
Explanation:
As we know that,
Molarity = $/dfrac{moles}{volume}$
Moles = $/dfrac{mass}{molar mass}$
Given,
mass of Na2CO3 =25.3g
Molar mass of Na2CO3 = 106g
Vol of sol.=250mL=0.25L
moles of Na2CO3 = $ /dfrac{25.3}{106} $ = 0.2386
Molarity of Na2CO3 = $/dfrac{0.2386}{0.25}$ = 0.955M
When Na2CO3 dissociates, it gives 2 moles of Na+ ion and 1 mole of CO32-
∴ Molar concentration of sodium ion,(Na+ )=2×0.955=1.910M
Molar concentration of carbonate ion, (CO32-) =1×0.955=0.955M
54211.10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount j of
water produced in this reaction will be
water produced in this reaction will be
3 mol
4 mol
1 mol
2 mol
Explanation:
First we see reaction by which form water
2H2+ O2--->2H2O
here we see 2mole hydrogen react with 1mole Oxygen form two mole water
we have
10gm hydrogen =5mole hydrogen
64gm Oxygen =2mol mole oxygen
we know according to above reaction
4gm hydrogen react with 32gm oxygen
so,
10gm hydrogen react with 80gm oxygen
but we have given amount of oxygen only 64 gm it means Oxygen is here limiting reagent .
now
all oxygen reat then 8gm hydrogen react and form 4mole of water .
First we see reaction by which form water
2H2+ O2--->2H2O
here we see 2mole hydrogen react with 1mole Oxygen form two mole water
we have
10gm hydrogen =5mole hydrogen
64gm Oxygen =2mol mole oxygen
we know according to above reaction
4gm hydrogen react with 32gm oxygen
so,
10gm hydrogen react with 80gm oxygen
but we have given amount of oxygen only 64 gm it means Oxygen is here limiting reagent .
now
all oxygen reat then 8gm hydrogen react and form 4mole of water .
54212.What volume of oxygen gas (O2 ) measured at 0°C and 1 atm, is needed to bum completely
l L of propane gas (C3H8 ) measured under the same conditions?
l L of propane gas (C3H8 ) measured under the same conditions?
5 L
10 L
7 L
6 L
Explanation:
According to the above equation1 vol. or 1 liter of propane requires to 5 vol. or 5 liter of O2 to burn completely.
According to the above equation1 vol. or 1 liter of propane requires to 5 vol. or 5 liter of O2 to burn completely.
54213.How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO
and 3.2 g HCl?
and 3.2 g HCl?
0.011
0.029
0.044
0.333
Explanation:
The equation for the reaction between PbO and HCl is as follows:
PbO + 2HCl------------> PbCl2 + Cl + H2O
Therefore 1 moles of PbO require 2 moles of HCl to give 1 mole of PbCl2.
Calculate the reagent, between HCl and PbO, that is the limiting reagent.
Pb-207.2 Cl-35.5 O-16 H-1
moles=mass/molar mass
moles of PbO = 6.5/(207+16) = 6.5/223.2 = 0.029148
if 1 mole = 2 moles of HCl
then 0.029148= 2x0.029148= 0.058296
moles of HCl= 3.2/(1+35.5) = 3.2/36.5 = 0.08767
if 2 moles of HCl= 1 mole of PbO
then 0.08767= 1x 0.8767/2 = 0.04384
Therefore PbO is the limiting reagent.
Calculate moles of PbCl2 produced as follows:
if 1 mole of PbO = 1 mole of PbCl2
then 0.029148 will give= 1 x 0.029148 moles
Therefore moles of PbCl2 produced = 0.029148 moles
The equation for the reaction between PbO and HCl is as follows:
PbO + 2HCl------------> PbCl2 + Cl + H2O
Therefore 1 moles of PbO require 2 moles of HCl to give 1 mole of PbCl2.
Calculate the reagent, between HCl and PbO, that is the limiting reagent.
Pb-207.2 Cl-35.5 O-16 H-1
moles=mass/molar mass
moles of PbO = 6.5/(207+16) = 6.5/223.2 = 0.029148
if 1 mole = 2 moles of HCl
then 0.029148= 2x0.029148= 0.058296
moles of HCl= 3.2/(1+35.5) = 3.2/36.5 = 0.08767
if 2 moles of HCl= 1 mole of PbO
then 0.08767= 1x 0.8767/2 = 0.04384
Therefore PbO is the limiting reagent.
Calculate moles of PbCl2 produced as follows:
if 1 mole of PbO = 1 mole of PbCl2
then 0.029148 will give= 1 x 0.029148 moles
Therefore moles of PbCl2 produced = 0.029148 moles
54214.An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave
C, 38.71% and H, 9.67%. The empirical formula of the compound would be
C, 38.71% and H, 9.67%. The empirical formula of the compound would be
CHO
CH4O
CH3O
CH2O
Explanation:
The empirical formula is to be calculated as:
Element - C
% Composition - 38.71
Atomic Mass - 12
Mole Ratio - (38.7/12) = 3.22
Simple Ratio - (3.22/3.22) =1
Element - H
% Composition - 9.67
Atomic Mass - 1
Mole Ratio - (9.67/1) = 9.67
Simple Ratio - (9.67/3.22) = 3
Element - O
% Composition - 100-(38.71+9.67) =51.62
Atomic Mass - 16
Mole Ratio - (51.62/16) = 3.22
Simple Ratio - (3.22/3.22) =1
Hence the empirical formula of the compound is CH3O
The empirical formula is to be calculated as:
Element - C
% Composition - 38.71
Atomic Mass - 12
Mole Ratio - (38.7/12) = 3.22
Simple Ratio - (3.22/3.22) =1
Element - H
% Composition - 9.67
Atomic Mass - 1
Mole Ratio - (9.67/1) = 9.67
Simple Ratio - (9.67/3.22) = 3
Element - O
% Composition - 100-(38.71+9.67) =51.62
Atomic Mass - 16
Mole Ratio - (51.62/16) = 3.22
Simple Ratio - (3.22/3.22) =1
Hence the empirical formula of the compound is CH3O
54215.An element, X has the following isotopic composition:
The weighted average atomic mass of the naturally occurring element X is closest to
The weighted average atomic mass of the naturally occurring element X is closest to
201 amu
202 amu
199 amu
200 amu
Explanation:
Formula:-
Average atomic mass= [Atomicmassof X 200 × abundance + Atomicmassof X 199 × abundance + Atomicmassof X 202 × abundance]/100
=$/dfrac{200×90+199×8.0+202×2.0}{100}$
=$/dfrac{18000+1592+404}{100}$
=199.96≈200amu.
Formula:-
Average atomic mass= [Atomicmassof X 200 × abundance + Atomicmassof X 199 × abundance + Atomicmassof X 202 × abundance]/100
=$/dfrac{200×90+199×8.0+202×2.0}{100}$
=$/dfrac{18000+1592+404}{100}$
=199.96≈200amu.
54216.The maximum number of molecules is present in
15L of H2 gas at STP
5 L of N2 gas at STP
0.5 g of H2 gas
10 g of O2 gas.
Explanation:
option 1)
The maximum number of molecules is present in 15 L of H2 gas at STP.
1 mole of a gas at STP occupies a volume of 22.4 L.
15 L of gas corresponds to $/dfrac{15L}{22.4L}$ = 0.67 moles.
option 2)
5L of N2 gas at STP :
1 mole of a gas at STP occupies a volume of 22.4 L.
5 L of gas corresponds to $/dfrac{5L}{22.4L}$ = 0.22 moles.
option 3)
0.5 g of H2 gas (molecular weight 2g/mol) corresponds to $/dfrac{0.5g}{2g/mol}$ = 0.25 moles
option 4)
10 g of O2 gas (molecular weight 3 g/mol) corresponds to $/dfrac{10g}{32g/mol}$ = 0.3125 moles
Higher is the number of moles of a gas, higher will be its number of molecules.
Therefore, the correct option is A.
option 1)
The maximum number of molecules is present in 15 L of H2 gas at STP.
1 mole of a gas at STP occupies a volume of 22.4 L.
15 L of gas corresponds to $/dfrac{15L}{22.4L}$ = 0.67 moles.
option 2)
5L of N2 gas at STP :
1 mole of a gas at STP occupies a volume of 22.4 L.
5 L of gas corresponds to $/dfrac{5L}{22.4L}$ = 0.22 moles.
option 3)
0.5 g of H2 gas (molecular weight 2g/mol) corresponds to $/dfrac{0.5g}{2g/mol}$ = 0.25 moles
option 4)
10 g of O2 gas (molecular weight 3 g/mol) corresponds to $/dfrac{10g}{32g/mol}$ = 0.3125 moles
Higher is the number of moles of a gas, higher will be its number of molecules.
Therefore, the correct option is A.
54217.Which has maximum molecules?
7g N2
2g H2
16g NO2
16g O2
Explanation:
No. of moles = $\dfrac{Weight (W)}{Molecular weight (M)}$
Therefore,
(a) No. of moles = $\dfrac{7}{28}$ = 0.25 moles
(b) No. of moles = $\dfrac{2}{2}$ = 1 moles
(c) No. of moles = $\dfrac{16}{46}$ = 0.347 moles
(d) No. of moles = $\dfrac{16}{32}$ = 0.5 moles
Since H2 has maximum no. of moles, hence it has maximum no. of molecules.
No. of moles = $\dfrac{Weight (W)}{Molecular weight (M)}$
Therefore,
(a) No. of moles = $\dfrac{7}{28}$ = 0.25 moles
(b) No. of moles = $\dfrac{2}{2}$ = 1 moles
(c) No. of moles = $\dfrac{16}{46}$ = 0.347 moles
(d) No. of moles = $\dfrac{16}{32}$ = 0.5 moles
Since H2 has maximum no. of moles, hence it has maximum no. of molecules.
54218.Embryo sac is to ovule as.........is to an anther.
stamen
filament
pollen grain
androecium
Explanation:
Embryo sac is present in ovule, while pollen grains are present in anther.
Embryo sac is present in ovule, while pollen grains are present in anther.
54219.Choose the correct statement from the following:
Cleistogamous flowers always exhibit autogamy.
Chasmogamous flowers always exhibit geitonogamy.
Cleistogamous flowers exhibit both autogamy and geitonogamy.
Chasmogamous flowers never exhibit autogamy.
Explanation:
Chasmogamous flowers exhibit both autogamy(self-pollination)and allogamy (cross-pollination). While, in cleistogamous flower, the anthers and stigma lie close to each other with in the closed flowers. When anthers dehisces in the flower buds, pollen grains come in contact with the stigma for effective pollination. Thus, these flowers are invariably autogamous as there is no chance of cross-pollen landing on the stigma.
Chasmogamous flowers exhibit both autogamy(self-pollination)and allogamy (cross-pollination). While, in cleistogamous flower, the anthers and stigma lie close to each other with in the closed flowers. When anthers dehisces in the flower buds, pollen grains come in contact with the stigma for effective pollination. Thus, these flowers are invariably autogamous as there is no chance of cross-pollen landing on the stigma.
54220.While planning for an artificial hybridisation programme involving dioecious plants, which of the
following steps would not be relevant?
following steps would not be relevant?
Bagging of female flower.
Dusting of pollen on stigma
Emasculation
Collection of pollen
Explanation:
Emasculation is the process of removal of the anthers of a flower in order to prevent self-pollination or the undesirable pollination of neighbouring plants.In case of dioecious plants where male and female flowers do not grow at the same plant,there is no need of emasculation.
Emasculation is the process of removal of the anthers of a flower in order to prevent self-pollination or the undesirable pollination of neighbouring plants.In case of dioecious plants where male and female flowers do not grow at the same plant,there is no need of emasculation.
54221.In a typical complete, bisexual and hypogynous flower the arrangement of floral whorls on the thalamus from the outermost to the innermost is
calyx, corolla, androecium and gynoecium
calyx, corolla, gynoecium and androecium
gynoecium, androecium, corolla and calyx
androecium, gynoecium, corolla and calyx
Explanation:
The calyx, a whorl of sepals (outermost).
The corolla, a whorl of petals (inside the calyx).
The androecium, a whorl of stamens (inside the corolla).
The - gynoecium, a whorl of pistils ( forms the inner most whorl).
The calyx, a whorl of sepals (outermost).
The corolla, a whorl of petals (inside the calyx).
The androecium, a whorl of stamens (inside the corolla).
The - gynoecium, a whorl of pistils ( forms the inner most whorl).
54222.In a fertilised embryo sac, the haploid, diploid and triploid structures are
synergid, zygote and primary endosperm nucleus
synergid, antipodal and polar nuclei
antipodal, synergid andprimary endosperm nucleus
synergid, polar nuclei and zygote
Explanation:
Diploid secondary nucleus fertilises with a haploid male gamete to form a triploid PEN.
Diploid secondary nucleus fertilises with a haploid male gamete to form a triploid PEN.
54223.From among the situations given below, choose the one that prevents both autogamy and
geitonogamy.
geitonogamy.
Monoecious plant bearing unisexual flowers.
Dioecious plant bearing only male or female flowers.
Monoecious plant with bisexual flowers.
Dioecious plant with bisexual flowers.
Explanation:
Autogamy is a method of self-pollination in which the transfer of polien grains from anther to stigma of the same flower takes place. While geitonogamy, is the transfer of pollen grains from anther to stigma of another flower of the same plant.In the above condition, Autogamy can happen in case of bisexual flowers. Geitonogamy can happen in case of dioecious plants bearing only male or female flowers.
Autogamy is a method of self-pollination in which the transfer of polien grains from anther to stigma of the same flower takes place. While geitonogamy, is the transfer of pollen grains from anther to stigma of another flower of the same plant.In the above condition, Autogamy can happen in case of bisexual flowers. Geitonogamy can happen in case of dioecious plants bearing only male or female flowers.
54224.The phenomenon observed in some plants where in parts of the sexual apparatus is used for forming
embryos without fertilisation is called
embryos without fertilisation is called
parthenocarpy
apomixis
vegetative propagation
sexual reproduction
Explanation:
Apomixis is the phenomenon of formation of seeds without fertilisation.Parthenocarpy is the formation of fruits without fertilisation and hence the fruits are seedless,e.g., banana. Vegetative propagatian or reproduction is a form of asexual reproduction in plants, in which new organisms arise without
production of seeds or spores.Sexual reproduction involves formation of the male and female gametes, either by the same individual or by different individuals of the opposite sex. These gametes fuse of form the zygote which develops to form the new organism.
Apomixis is the phenomenon of formation of seeds without fertilisation.Parthenocarpy is the formation of fruits without fertilisation and hence the fruits are seedless,e.g., banana. Vegetative propagatian or reproduction is a form of asexual reproduction in plants, in which new organisms arise without
production of seeds or spores.Sexual reproduction involves formation of the male and female gametes, either by the same individual or by different individuals of the opposite sex. These gametes fuse of form the zygote which develops to form the new organism.
54225.Among the terms listed below, those that of are not technically correct names for a floral whorl are
(i) androecium
(ii) carpel
(iii) corolla
(iv) sepal
(i) androecium
(ii) carpel
(iii) corolla
(iv) sepal
(i)and(iv)
(iii) and (iv)
(ii) and (iv)
(i) and (ii)
Explanation:
Sepals collectively form a whorl, called as calyx while technically the carpel is known as gynoecium.
Corolla and androecium are the names given to petals and stamens respectively.
Sepals collectively form a whorl, called as calyx while technically the carpel is known as gynoecium.
Corolla and androecium are the names given to petals and stamens respectively.
54226.In a flower, if the megaspore mother cell forms megaspores without undergoing meiosis and if one of
the megaspores develops into an embryo sac, its nuclei would be
the megaspores develops into an embryo sac, its nuclei would be
haploid
diploid
a few haploid and a few diploid
with varying ploidy
Explanation:
Formation of megaspores includes asexual reproduction which occurs in the absence of pollinators . It occurs in the
megaspore mother cell without undergoing meiosis, and produces diploid embryo sac through mitotic divisions.
Haploid cells will be formed during sexual reproduction when cell will undergo meiosis and option ‘c’ and ‘d’
is not shown by megaspore mother cell.
Formation of megaspores includes asexual reproduction which occurs in the absence of pollinators . It occurs in the
megaspore mother cell without undergoing meiosis, and produces diploid embryo sac through mitotic divisions.
Haploid cells will be formed during sexual reproduction when cell will undergo meiosis and option ‘c’ and ‘d’
is not shown by megaspore mother cell.
54227.In an embryo sac, the cells that degenerate after fertilisation are
synergids and primary endosperm cell
synergids and antipodals
antipodals and primary endosperm cell
egg and antipodals
Explanation:
Primary endosperm cell provides food for the growing embryo, while egg develops into embryo.On the other hand,in fertilised embryo sac antipodals and synergids gradually degenerate after the formation of zygote.
Primary endosperm cell provides food for the growing embryo, while egg develops into embryo.On the other hand,in fertilised embryo sac antipodals and synergids gradually degenerate after the formation of zygote.