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Sonja [21]
3 years ago
6

You need to make an aqueous solution of 0.150 M sodium fluoride for an experiment in lab, using a 500 mL volumetric flask. How m

uch solid sodium fluoride should you add?
Chemistry
1 answer:
storchak [24]3 years ago
7 0

Answer:

Explanation:

Solution required is .15 M 500 mL sodium fluoride .

molecular weight of sodium fluoride = 42

one mole of sodium fluoride = 42 grams

No of moles contained in the solution of .15M , 500 mL solution

= .15 x 0.5 moles = .075 moles

1 mole = 42 g

.075 moles = .075 x 42 = 3.15 grams .

So mass of sodium fluoride required = 3.15 g .

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A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
tigry1 [53]

Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

6 0
3 years ago
To measure the volume of the blood system of an animal, the following experiment was done. A 1.0-mL sample of an aqueous solutio
Artyom0805 [142]

Answer:199mL

Explanation:

Let V ml be the volume of blood in animal.

When 1.0 ml sample is added, total volume becomes V+1.0ml. Its activity is 1000 dpm.

After equilibrium, 2.0 ml of the sample had activity of 10 dpm.

Hence, after equilibrium, the activity of V+1.0 ml of blood sample will be 10/2 (V+1.0ml)=1000dpm

Hence, V=199ml.

4 0
3 years ago
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