5. What is the molarity (M) of a solution that results when 30.2g of (NH4)3PO4 d

Question

5. What is the molarity (M) of a solution that results when 30.2g of (NH4)3PO4 dissolved in H2O and diluted to exactly 250.0 mL? 6. Which compound is the oxidizing agent in the reaction below? 7. An aqueous hydrochloric acid solution has a pH of 3.15. What mass of HCl is present in 1.0 L of this solution? 8. How many moles of NaOH are present in 25 mL of a 0.1000M NaOH solution? 9. If you have 50.0 mL of 3.0 NaOH and you want 0.50 M NaOH. How much water in mL do you need to make the 0.50 M NaOH solution? 10. What is the oxidation number of P in a compound, H3PO4?

Explanation / Answer

5) b
molarity=30.2 /149 x 1000/250 =0.0852 mol/l = 0.811

6) d

7) pH = -log[H+]
[H+] = 10^-pH
[H+] = 10^-pH
[H+] = 10^-3.15
[H+] = 7.079x10^-4 mol/L

So there are 7.079x10^-4 moles H+ in 1L of solution
Since 1 mole HCl has 1 mol H+ then the concentration of HCl is also
7.079x10^-4 mol/L

Si in 1 L
mass HCl = molar mass x moles
= 36.458 g/mol x 7.079x10^-4 moles
= 2.58x10^-2 g
= 2.6x10^-2 g (2 sig figs)

Ans is a

8) Number of Moles = Concentration * volume in liters.

=0.1*0.025

0.0025

=2.5*10^-3

Ans is b

9) just dilution principal M1V1=M2V2

50*3=0.5*V2

V2 = 300mL

Ans is d

10) O is -2 and H is +1. The total of the oxidation numbers has to add up to the charge on the entity, which is 0.

3H + P + 4O = 0
3(1) + P + 4(-2) = 0
3+P-8 = 0
P = +5

Ans is b

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