Question 16 Of the following substances, ________ has the highest boiling point.

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.

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