Answered

A student dissolved 3.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are in the final solution?

Answer :

Answer:

0.081g

Explanation:

first step:calculate the molar conc of the stock solution

molar conc=mass/molar mass ×1000/vol(ml)

molar mass of Co(N03)2=182.943

molar conc=3g/182.943 ×1000/100ml

=0.163M

To calculate the conc of the resulting solution we apply the dilution principle

n=cv

C1V1=C2V2

C1=conc of stock solution=0.163M

V1=4mL

V2=275ml(final volume)

by making 'C2" the subject C2=0.163×4/275

=0.0024M

by applying the formular for calculating molar conc of a solution,we can calculate the mass of CO(NO3)2 in the reulting solution

molar conc=mass/molar mass ×1000(L)/vol(mL)

mass=0.0024×182.943×275/1000

=0.1193g

Mass of CO(NO3)2  in the resulting solution=0.1199g

since CO+(NO3)2=CO(NO3)2

Percentage by mass of (NO3)2 in CO(NO3)2=(NO3)2/CO(NO3)2

=124/182.943×100

=67.8%

it means 67.8% of 0.1193g is the mass of (NO3)2=0.678×0.1199

=0.081g

pstnonsonjoku

The mass of NO3 ion present in the final solution  is 0.14 g.

Molar mass of Co(NO3)2 = 183 g/mol

Number of moles of Co(NO3)2 =  3.00 g/183 g/mol = 0.016 moles

Concentration of  Co(NO3)2 = number of moles / volume

Concentration of  Co(NO3)2 = 0.016 moles /0.1 L = 0.16 M

From the dilution formula;

C1V1 =C2V2

C1 = 0.16 M

C2 = ?

V1 =  4.00 mL

V2 = 275. mL

C2 = C1V1 /V2

C2 = 0.16 M × 4.00 mL/275. mL

C2 = 0.0023 M

Molar mass of NO3 ion = 62 g/mol

Mass of NO3 ion = 0.0023 M × 62 g/mol

Mass of NO3 ion = 0.14 g

The mass of NO3 ion present in the final solution is  0.14 g.

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