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Chemistry : asked on vskalaniedits
 31.12.2020

1.18 g of silver nitrate (AgNO3) are dissolved in 25.0 mL of water (ρ = 1,000 g L-1). Calculate (to three decimal places) the molality (m, mol kg-1) of the solution.

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13.11.2022, solved by verified expert
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Ranjana Bhatt
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Answer:

Milality of silver nitrate

Step-by-step explanation:

Mass of Silver nitrate = 1.18 g

Molar mass of Ag NO3 = 170g( Ag=108g, N=14g, O=16g)

Density of water = 1000g/cm3 (given in question)

Volume of water (solvent) =25ml

Mass =volume × density

           25×1000 = 25000 gram

        = 25 ky of solvent

Molality (m) = moles/mass of solvent in my

Moles of AgNO3 =1.18/170

m = 1.18/ 170×25

    = 0.000277

   =2.8 × 10^-4 m

       

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