16.7 g of silver nitrate is dissolved in 98.32 g of water for a total solution m

Question

16.7 g of silver nitrate is dissolved in 98.32 g of water for a total solution mass of 101.16 g that is at 4.0 degree C. Molarity of silver nitrate? Molarity of silver? Molarity of silver and silver nitrate? Mass % silver? | (solution B) 0.327 mL of the above solution is diluted to 1.00 L also at 4.0 degree C. What is the new molarity of silver? What is the mass %o silver? How many ppm of silver are in this solution? Express concentration of silver nitrate in g/mL. 10.0000 mL of the above solution is placed in a vial and allowed to warm to 40 degree C in a sealed container. Has the molarity of silver in the solution changed? If so, what is the new value? (solution C) Suggest a dilution scheme to make 500. mL of a 0.357 ppb solution of silver at 20 degree C using one of the above solutions.

Explanation / Answer

Molarity = Moles of Solute / Volume of solution in L

Molarity of Silver nitrate = Mass of Silver nitrate / MW of Silver nitrate * Volume of solutiom

Assuming density of water and density of AgNO3 are same = 1gm /ml

Mass of solution = Volume of solution = 101.16 ml

Molarity of AgNO3 = 0.97 M

Molality = Moles of solute / Mass of solvent in Kg

MOlality of AgNO3 = 16.7*1000 / 170*98.32 = 0.999

In 1 mole of AgNO3 there are 1:1 moles of Ag and NO3

So Molarity of Ag = Molarity of AgNO3 = 0.97 M

Since Moles of Ag = Moles of AgNO3

Molality of Ag = 0.999

Mass % of Ag = MOles of Silver * MW*100 / Mass of AgNO3

Mass % of Ag = 0.0982 *107.8682 *100 / 16.7 = 63.42 %

Part B

MOles of Silver = Molarity of silver*Volume in L = 0.97*0.327 / 1000 = 3.1719*10^-4

Molarity of Silver = Moles of silver / V in L = 3.1719 *10^-4 M

Mass % of silver = Mass of silver *100 / Mass of AgNO3 = 0.2 %

1 ppm = 1 mg solute per liter of solution

3.1719*10^-4 M = 34.21 ppm of Silver

Molarity to gm /ml

3.1719*10^-4 M ==> 3.42*10^-5 gm /ml

ON increasing Temperature ; Molarity decreases

V1 / V2 = T1 / T2

10 / V2 = 273+4 / 273+40

V2 = 11.3 ml

New Molarity = ?

11.3ml *3.42*10^-5 gm /ml = 3.8644*10^-4 gms

Molarity = 3.17 *10^-4 M ---> New One

1 ppb means

and 1 ppb = 0.001 ppm

1g per 10^9 gms of solution

0.357 ppb = 0.000357 ppm = 0.000357 mg / L

By using solution B we can prepare the 500 ml of 0.357 ppb by dilutions

Using Dilution formula

M1V1 = M2V2

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