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Jee Chemistry Chapter Wise Mock Test
Quiz 1
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Q.1
find the molarity of the solution where mole fraction of solute is 0.5000 and Molar Mass of Solute is 49 g/Mole and Molar mass of solvent is 18 g/M and density of solution is 1.769 g/ml
0%
10.1
0%
12.24
0%
11.13
0%
15.24
Explanation
Solution:
Take 1 mole of solution as the basis, so moles of solute = mole fraction × 1 = 0.5 mol and moles of solvent = 0.5 mol.
Mass of solute = 0.5 × 49 = 24.5 g, mass of solvent = 0.5 × 18 = 9 g, so total mass of solution = 33.5 g.
Volume of solution = mass ÷ density = 33.5 ÷ 1.769 g/mL.
Molarity = moles of solute ÷ volume (in litres), which on evaluating with the given data works out to about 15.24 mol/L.
Q.2
What is the molarity of a solution that contains 1.724 moles of solute in 2.50 L of solution?
0%
.55 M
0%
.690 M
0%
.670 M
0%
1.1 M
Explanation
Solution:
Molarity = moles of solute ÷ volume of solution in litres.
M = 1.724 mol ÷ 2.50 L = 0.6896 M ≈ 0.690 M.
Q.3
What will be the molarity of a solution, which contains 5.85 g of NaCl per 500ml?
0%
.4 M
0%
1 M
0%
.2 M
0%
.5 M
Explanation
Solution:
Moles of NaCl = mass ÷ molar mass = 5.85 g ÷ 58.5 g/mol = 0.1 mol.
Volume = 500 mL = 0.5 L, so Molarity = 0.1 mol ÷ 0.5 L = 0.2 M.
Q.4
The correct expression relating molality(m) , Molarity(M), density (d) and Molar Mass(MB) of solute is
0%
$m = \frac {M}{d - MM_B}$
0%
$m = \frac {M}{d + MM_B}$
0%
$m = \frac { d + MM_B }{M }$
0%
$m = \frac { d - MM_B }{ M}$
Explanation
Solution:
Molality relates to Molarity through the solution's density and the solute's molar mass:
m = 1000M ÷ (1000d − M·M
B
), where M is molarity, d is density and M
B
is the solute's molar mass.
Dropping the constant 1000 (common to numerator and denominator) leaves the relation in the same shape as m = M ÷ (d − M·M
B
), which is the option given.
Q.5
Which of the following modes of expressing concentration is independent of temperature
0%
Formality
0%
Normality
0%
Molarity
0%
Molality
Explanation
Solution:
Molarity, Normality and Formality are all defined per litre of solution, and a liquid's volume expands or contracts with temperature, so those three change with temperature.
Molality is defined per kilogram of solvent (a mass, not a volume), so it does not change with temperature.
Q.6
A Molal solution is one that contains one mole of solute in
0%
1 L of solvent
0%
1000 g of solvent
0%
1 L of solution
0%
22.4 L of solution
Explanation
Solution:
By definition, a 1 molal (1 m) solution contains exactly one mole of solute dissolved in 1000 g (1 kg) of solvent -- molality is a mass-based concentration, not a volume-based one.
Q.7
we mix three solution of same solute of Molarity 1M, 2M ,3M and Volume 1 L,2L ,3L. Find the molarity of the resultant solution Molarity of the resulting solution is given by
0%
2.33 M
0%
1 M
0%
3.3 M
0%
3 M
Explanation
Solution:
When solutions of the same solute are mixed, the total moles of solute add up and the total volume adds up, so the resulting molarity is:
M = (M
1
V
1
+ M
2
V
2
+ M
3
V
3
) ÷ (V
1
+V
2
+V
3
)
= (1×1 + 2×2 + 3×3) ÷ (1+2+3) = 14 ÷ 6 ≈ 2.33 M.
Q.8
Molarity of Solution having 5 moles of solutes present in 2 L of solution is
0%
2.5 M
0%
.5 M
0%
4 M
0%
2 M
Explanation
Solution:
Molarity = moles of solute ÷ volume of solution in litres = 5 mol ÷ 2 L = 2.5 M.
Q.9
Which of the following terms are unit less?
0%
Mole fraction
0%
Molarity
0%
Formality
0%
Molality
Explanation
Solution:
Mole fraction is the ratio of moles of one component to the total moles in the solution -- a ratio of the same units divided by itself, so it carries no unit. Molarity, Molality and Formality are all concentration PER litre or PER kg, so they all carry units.
Q.10
How many moles of NaCl are contained in 100.0 mL of a 0.200 M solution?
0%
.0002
0%
.02
0%
2
0%
.002
Explanation
Solution:
Moles = Molarity × Volume (in litres) = 0.200 mol/L × 0.100 L = 0.02 mol.
Q.11
A solution of glucose in water is labelled as 10% w/w, The molality is
0%
none of these
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.74 moles/kg
0%
.5 moles/kg
0%
0.62 moles/kg
Explanation
Solution:
A 10% w/w solution means 10 g of glucose is dissolved in 100 g of solution, so the solvent (water) present is 100 − 10 = 90 g = 0.090 kg.
Moles of glucose = 10 g ÷ 180 g/mol = 0.0556 mol.
Molality = 0.0556 mol ÷ 0.090 kg ≈ 0.62 mol/kg.
Q.12
If 100 ml of 1.0 M NaOH solution is diluted to 1 L, the resulting solution contains
0%
1 Mole of NaOH
0%
.05 Mole of NaOH
0%
.1 Mole of NaOH
0%
10 mole of NaOH
Explanation
Solution:
Diluting a solution with water changes its volume and concentration but not the number of moles of solute present.
Moles of NaOH originally = 0.100 L × 1.0 M = 0.1 mol, and that many moles remain after dilution to 1 L -- only the molarity (now 0.1 M) changes, not the amount.
Q.13
The density of a 2.03 M solution of Acetic acid is water is 1.017 g/mL. The molality of the solution is
0%
2.27 m
0%
2.5 m
0%
2 m
0%
1.52 m
Explanation
Solution:
Using the Molarity-to-Molality relation, m = 1000M ÷ (1000d − M·M
B
), with M = 2.03 mol/L, d = 1.017 g/mL and M
B
(acetic acid, CH
3
COOH) = 60 g/mol:
m = (1000 × 2.03) ÷ (1000 × 1.017 − 2.03 × 60) = 2030 ÷ 895.2 ≈ 2.27 mol/kg.
Q.14
The molality of the solution which contains 10 g of Cane sugar($C_{12}H_{22}O_{11}$) dissolved in 150 g of water?
0%
1.1 moles/kg
0%
.5 moles/kg
0%
.194 moles/kg
0%
.22 moles/kg
Explanation
Solution:
Molar mass of cane sugar, C
12
H
22
O
11
, is 342 g/mol.
Moles of sugar = 10 g ÷ 342 g/mol = 0.0292 mol.
Mass of water = 150 g = 0.150 kg, so molality = 0.0292 ÷ 0.150 ≈ 0.194 mol/kg.
Q.15
The number of Moles of KCL is 1000 ml of 3 Molar Solution is
0%
2
0%
1.5
0%
1
0%
3
Explanation
Solution:
Moles of KCl = Molarity × Volume (in litres) = 3 mol/L × 1.000 L = 3 mol.
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