Question

Quiz 8 X What is the molarity of 0.945 L of a solution that contains 0.227 moles of Fe(NO3)3? M 0.227 nol O.945L 응급주 240 M O.227 0.445 24.02M 240 What is the molarity of a solution that contains 50.0 grams of sodium hydroxide, NaOH, in 855 mL of solution? The molar mass of NaOH is 40.0 grams. imoi 1,25 moi No DH 509 409 S0 1585 M M: .SSL n 1,25 T55 mL L 3 What mass of zinc nitrate, Zn(NO3)2, is needed to prepare 675 mL of 1.20 M zinc nitrate 75ML- L aqueous solution? 7.20 18 Co.75L 4) Hydrochloric acid is obtained commercially at a concentration of 12.1 M. How many milliliters of 12.1 M HCI(aq) must be used to prepare 2.00 x 103 mL of 0.500M HCl(aq)? 12. 12.1F6.S00 2.00x10 ML = 200 *ree 2000L 2000 .3mL 5) How many milliliters of 18.0 M sulfuric acid, H2SO04(aq)' must be diluted with water to prepare 5.00 x 103 mL of 0.250 M sulfuric acid? I8a5 VI 365 ML 6) What is the molarity of a solution in which 10.8 grams of calcium chloride is dissolved in enough water to make 125 mL of solution? 10.34 M: X100 8lo4 M 1.25L If 37.6 grams of sodium bromide, NaBr, are added to enough water to make a solution whose molarity is 0.400 M, how many liters of solution do you have? L02.8 0.460M 2.q 25700 mL X 1o0 25700-L 257L

Answer 1

Answer 1 - Given,

moles of Fe(NO₃)₃ = 0.227 mol

Volume = 0.945 L

Molarity = ?

We know that,

Molarity = Moles/ Volume in L

Molarity = 0.227mol/0.945L

Molarity = 0.240 M [Answer]

Answer 2 -

Given,

Mass of NaOH = 50.0 g

Volume = 855 ml or 0.855 L

Molar Mass of NaOH = 40 g/mol

Molarity = ?

We know that,

Molarity = Moles/ Volume in L

Moles = Mass/Molar Mass

So,

Molarity = Mass/(Molar mass* volume in L)

Molarity = 50.0 g/(40 g/mol * 0.855 L)

Molarity = 1.46 M [Answer]

Answer 3-

Given,

Volume = 675 ml or 0.675 L

Molarity = 1.20 M

Molar Mass of Zn(NO₃)₂ = 189.36 g/mol

mass = ?

We know that,

Molarity = Moles/ Volume in L

Moles = Mass/Molar Mass

So,

Molarity = Mass/(Molar mass* volume in L)

Mass = Molarity *(Molar mass* volume in L)

Put the values,

Mass = 1.20 M *(189.36 g/mol* 0.675 L)

Mass = 153.3816 g [Answer]

Answer 4 -

Given,

Initial Molarity of HCL = 12.1 M

Final Molarity of HCl = 0.500 M

Final Volume = 2.00 * 10³ ml or 2.00 L [1 ml = 0.001 L]

Initial Volume = ?

We know that,

M1V1 = M2V2

where, M1 and M2 are Molarity

V1 and V2 are Volumes

Put the values,

12.1 M* V1 = 0.500 M * 2.00 L

V1 = (0.500 M * 2.00 L)/12.1M

V1 = 0.083 L or 83 ml [1 L = 1000 ml]

So, 83 ml of 12.1 M HCl is required. [Answer]