The average atomic mass of chlorine using the following data would be -
% Natural
Abundance
Molar
Mass
38.4527 u
37.4527 u
36.4527 u
35.4527 u
The number of moles of hydrogen atoms present in 3 moles of ethane (C2H6), will be-
18.069 x 1023
1.8 x 1023
18
6 x 1023
20 g of sugar is dissolved in enough water to make a final volume of 2L. The concentration of sugar (C12H22O11) in mol L–1 will be
2. 0.029 mol/L
0.29 mol/L
4. 0.032 mol/L
0.35 mol/L
If the density of methanol is 0.793 kg L–1, the volume needed for making 2.5 L of its 0.25 M solution would be
2. 24.78 mL
22.25 mL
4. 252 mL
25.22 mL
The level of contamination in water is 15 ppm (by mass of chloroform). Chloroform in the water sample would have a molality of
3.25 x 10-4
1.5 x 10-3
7.5 x 10-3
1.25 x 10-4
If the speed of light is 3.0 × 108 m s–1, then the distance covered by light in 2.00 nanoseconds will be -
0.500 m
0.600 m
0.700 m
0.800 m
In a reaction A + B2 →AB2, A will act as a limiting reagent if i. 300 atoms of A reacts with 200 molecules of Bii. 2 mol A reacts with 3 mol Biii. 100 atoms of A reacts with 100 molecules of Biv. 5 mol A reacts with 2.5 mol Bv. 2.5 mol A reacts with 5 mol BChoose the correct option
If 10 volumes of H2 gas react with 5 volumes of O2 gas, the volumes of water vapor produced would be
9
8
10
11
15.15 pm in the basic unit will be
1.515 x 10-12 m
2.57 x 10-11 m
2.87 x 10-11 m
1.515 x 10-11 m
Which will have the largest number of atoms among the following.
1 g Au (s)
1 g Na (s)
1 g Li (s)
1 g of Cl2(g)
The molarity of a solution of ethanol in water having a mole fraction of ethanol is 0.040 (assume the density of water to be 1) would be -
2.143 M
2.314 M
2.413 M
2.141 M
The mass of one 12C atom in g is -
1.393 x 10-23
1.993 x 10-23
1.773 x 10-23
1.593 x 10-23
The number of significant figures present in the answer of the following calculations [ (i), (ii), (iii)] are respectively -i. 0.02856 × 298.15 ×0.1120.5785ii. 5 × 5.364iii. 0.0125 + 0.7864 + 0.0215
The molar mass of naturally occurring Argon isotopes is
1. 49.99947 g mol-1
The highest number of atoms is present in -
(2) 52 u of He
52 moles of Ar
(3) 52 g of He.
All of the above have the same number of atoms
Burning a small sample of welding gas (constituted of C and H only) in oxygen gives 3.38 g carbon dioxide, 0.690 g of water, and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Molecular formula of welding gas would be -
C2H2
C2H6
CH
C2H4
The mass of CaCO3 required to react completely with 25 mL of 0.75 M HCl according to the given reaction would beCaCO3(s) + HCl(aq) ➡ CaCl2(aq) + CO2(g) + H2O(l)
0.36g
0.09 g
0.96 g
0.66 g
Amount of HCl that would react with 5.0 g of manganese dioxide, as per the given reaction will be4HCl(aq) + MnO2(s) ➡ 2H2O(l) + MnCl2(aq) + Cl2(2)
4.8 g
6.4 g
2.8 g
8.4 g
16 g of oxygen has the same number of molecules as in-
(a) 16 g of CO
(b) 28 g of N2
(c) 14 g of N2
(d) 1.0 g of H2
(a), (b)
(b), (c)
(c), (d)
(b), (d)
Unitless term among the following is/are-
(a) Molality
(b) Molarity
(c) Mole fraction
(d) Mass per cent
The correct match is
Which of the following figures does not represent 1 mole of dioxygen gas at STP?
(a) 16 g of gas
(b) 22.7 L of gas
(c) 6.022×1023 dioxygen molecules
(d) 11.2 L of gas
The correct choice among the above is -
(a, b, c)
(a, b, d)
(b, c, d)
(a, c, d)
The numbers 234,000 and 6.0012 can be represented in scientific notation as -
2.34×10-9 and 6×103
0.234 ×10-6 and 60012×10-9
2.34×10-9 and 6.0012×10-9
2.34×105 and 6.0012×100
The number of significant figures in the numbers 5005, 500.0, and 126,000 are, respectively:
2, 4, and 3
4, 1, and 3
4, 4, and 6
4, 4, and 3
0.50 mol Na2CO3 and 0.50 M Na2CO3 are different because -
Both have different amounts of Na2CO3
0.50 mol is the number of moles and 0.50 M is the molarity
0.50 mol Na2CO3 will generate more ions
None of the above
Round up the following number into three significant figures: i. 10.4107 ii. 0.04597 respectively are
The following data was obtained when dinitrogen and dioxygen react together to form different compounds:
The law of chemical combination applicable to the above experimental data is:
An organic compound contains 80% (by wt.) carbon and the remaining percentage of hydrogen. The right option for the empirical formula of this compound is: [Atomic wt. of C is 12, H is 1]
CH3
CH4
CH2
The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na Mol. Wt. 206. The maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin is-
1103
1206
2309
1412
SI unit of Mole fraction is -
Candela
Nm–1
kg
Unitless
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