Question 16 Of the following substances, ________ has the highest boiling point.
ID: 933285 • Letter: Q
Question
Question 16
Of the following substances, ________ has the highest boiling point.
H2O
SiH4
Ar
Cl2
BF3
2 points
Question 17
Gold has a density of 0.01932 kg/cm3. What is the mass (in kg) of a 92.5 cm3 sample of gold?
1.79
0.560
92.5
0.000209
4790
2 points
Question 18
Calculate the molality of a 10.0% (by mass) aqueous solution of hydrochloric acid.
3.32 m
121 m
1.10 m
1.66 m
The density of the solution is needed to solve the problem.
2 points
Question 19
A stock solution of HNO3 is prepared and found to contain 12.7 M of HNO3. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, the concentration of the diluted solution is ________ M.
0.254
1.57
0.635
635
254
2 points
Question 20
2.51
2.09
12.5
4.18
6.2
2 points
Question 21
0.0014
0.00086
120
0.0037
2.0
2 points
Question 22
13.5
28.8
54.1
47.3
94.6
2 points
Question 23
How many grams of H3PO4 are in 175 mL of a 4.00 M solution of H3PO4?
0.700
20.0
68.6
4.90
612
H2O
SiH4
Ar
Cl2
BF3
Explanation / Answer
Q:16: Among the given compounds SiH4, Cl2, Ar and BF3 are gases with very low boiling point. Hence water due to hydrogen bonding has the highest boiling point of 100 DegC. Hence H2O is the answer.
Q:17: Given the density of god, d = 0.01932 kg/cm3
The volume of the gold sample, V = 92.5 cm3
Hence mass of the given volume of gold sample, m = V x d = 92.5 cm3 x ( 0.01932 kg/cm3) = 1.79 Kg.
Hence the answer is 1.79 Kg (answer)
Q:18: Given the mass percentage of the aqueous solution = 10.0 %
=> [W(solute) / W(solution)]x100 = 10.0 %
=> W(solute) / W(solution) = 0.1
=> W(solute) / [W(solute) + W(solvent)] = 0.1
=> 1 + W(solvent) / W(solute) = 10
=> W(solvent) / W(solute) = 10-1 = 9
or W(solute) / W(solvent) = 1/9 = 0.111
The molecular mass of the solvent(H2O), M(solvent) = 18 g/mol
molecular mass of the solute (HCl), M(solute) = 36.5 g/mol
=> M(solute, Kg) = 0.0365 Kg/mol
Now molality can be calcualted as
m = moles of solute / W(solvent, in Kg) = W(solute, Kg) / [M(solute, Kg) x W(solvent, in Kg)]
=> m = [W(solute, Kg) / W(solvent, in Kg)] x [ 1 / M(solute, Kg)] = 0.111 x [1 / 0.0365 Kg/mol]
=> m = 3.32 m (answer)
Q:19: Given, M1 = 12.7 M, V1 = 25.0 mL = 0.025 L
M2 = ?, V2 = 0.500 L
Applying the law of dilution
M1V1 = M2V2
=> M2 = M1V1 / V2 = 12.7 M x 0.025 L / 0.500L = 0.635 M (answer)
Hence 3rd option is correct
Q:22: Given mass of ammonia(NH3, molar mass = 17.0 g/mol) = 14.4 g
Hence moles of NH3 = mass / molar mass = 14.4 g / 17.0 g/mol = 0.847 mol
The balanced chemical reaction is
4 NH3 (g) + 7 O2 (g) ------> 4 NO2 (g) + 6 H2O (g)
4 mole of NH3 reacts with 7 mol of O2.
Hence 0.847 mol of NH3 that will react with the moles of O2 = (7 mol O2 / 4 mol NH3) x 0.847 mol NH3
= 1.482 mol O2
Molecular mass of O2 = 32.0 g/mol
Hence mass of O2 = 1.482 mol x 32.0 g/mol = 47.3 g (answer)
Hence 4th option is correct.
Q:23: Given volume, V = 175 mL = 0.175 L
Concentration, M = 4.00 M
Henc moles of H3PO4 in the solution = MxV = 4.00 M x 0.175 L = 0.7 mol
Moleular mass of H3PO4 = 98.0 g/mol
Hence mass of H3PO4 in the solution = moles of H3PO4 x molecular mass = 0.7 mol x 98.0 g/mol
= 68.6 g (answer)
Hence 3rd option is correct.