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Question 1 1 pts

Calculate the molarity of 2.45 mol CaCl₂ in 0.3 L solution.

8.16 M

0.735 M

0.122 M

5.42 M

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Question 2 1 pts

What are the products of this acid-base reaction: HI(aq) + Ba(OH)₂(aq)?

BaI2 and H2O

BaH2 and I(OH)2

BaO and H2O

BaI2 and 2OH-

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Question 3 1 pts

Write the complete ionic equation for: Pb(NO₃)₂(aq) + 2KCl(aq) -->PbCl₂(s) + 2KNO₃(aq)

Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2Cl-(aq) -> PbCl2(s) + 2K+(aq) + 2NO3-(aq)

Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2Cl-(aq) -> Pb2+(aq) + 2NO3-(aq) + 2KCl(aq)

Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2Cl-(aq) -> Pb2+(aq) + 2Cl-(aq) + 2KNO3-(aq)

Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2Cl-(aq) -> Pb2+(aq) + 2Cl-(aq) + 2K+(aq) + 2NO3-(aq)

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Question 4 1 pts

What is the molarity of CO₃^2- in 0.24 M Al₂(CO₃)₃?

0.72 M

0.48

0.24 M

0.08 M

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Question 5 1 pts

If 430 mL of a 5.5 M glucose solution is diluted to 1500 mL, what is the molarity of the diluted solution?

1.58 M

2.0 M

0.13 M

3.2 M

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Question 6 1 pts

To what volume would you dilute 35 mL of a 9.5 M stock HCl solution to obtain a 1.4 M HCl solution?

238 mL

530 mL

300 mL

425 mL

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Question 7 1 pts

What volume of 2.5 M glucose solution contains 0.78 mols of glucose?

0.312 L

0.631 L

0.095 L

2.5 L

Answer 1

1. Moles of CaCL2 = 2.45

Volume of solution = 0.3 L

Molarity = Moles / Volume (L)

= 2.45 / 0.3

= 8.16 M

2. products of this acid-base reaction: HI(aq) + Ba(OH)2(aq) are -

2HI(aq) + Ba(OH)₂(aq)   2 H₂O(l) + BaI₂(aq)

3. Pb(NO3)2(aq) + 2KCl(aq)   PbCl2(s) + 2KNO3(aq)

The complete ionic equation for the reaction is -

Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2Cl-(aq) PbCl2(s) + 2K+(aq) + 2NO3-(aq)

4. molarity of CO32- in 0.24 M Al2(CO3)3 :

Concentration of Al₂ (CO3)₃ = 0.24 M

Moles of Al₂ (CO3)₃ in 1 L solution = 0.24 moles

Al₂ (CO3)₃   2Al³⁺   + 3CO3^2-

1 mole Al₂ (CO3)₃ on dissociation produce 3 moles of CO3^2- ions.

Therefore moles of CO3^2- in 1 L solution = 3 * 0.24

= 0.72 moles

Molarity of CO3^2- ions = 0.72 / 1

= 0.72 M

5. Initial volume of glucose solution, V1 = 430 mL

Initial molarity of solution, M1 = 5.5 M

Final volume of solution, V2 = 1500 mL

Using, M1*V1 = M2*V2

Final molarity of solution = M1*V1 / V2

= 5.5 * 430 / 1500

= 1.57 M

6. Initial volume of solution, V1 = 35 mL

Initial molarity, M1 = 9.5 M

Final Molarity, M2 = 1.4 M

Final Volume of solution, V2 = M1*V1 / M2

= 35 * 9.5 / 1.4

= 237.5 mL

7. Molarity of glucose solution = 2.5 M

Moles of glucose = 0.78

Molarity = Moles / Volume

Volume = 0.78 / 2.5

= 0.312 L

= 312.0 mL