Respuesta :

kenmyna
The mass  of  ni(NO3)2  that  dissolved  in  25.0 ml  of 0.100m  ni(NO3)2  solution  is  calculated  as   follows

fin  the  number  of  moles   =  molarity   x  volume in  liters

=25  x0.100/ 1000= 2.5  x10^-3  moles
mass  =  mass  x  molar  mass
= 2.5  x10^-3 moles x  182.71 g/mol  = 0.457  grams

Answer : The mass of nickel(II) nitrate is, 0.4568 grams

Explanation : Given,

Molar mass of nickel(II) nitrate = 182.71 g/mole

Volume of solution = 25.0 ml

Molarity = 0.100 M

Molarity : It is defined as the mass of solute present in one liter of solution.

Formula used :

[tex]Molarity=\frac{w\times 1000}{M\times V}[/tex]

where,

w = mass of nickel(II) nitrate

M = molar mass of nickel(II) nitrate

V = volume of solution

Now put all the given values in the above formula, we get:

[tex]0.100=\frac{w\times 1000}{182.71\times 25.0}[/tex]

[tex]w=0.4568g[/tex]

Therefore, the mass of nickel(II) nitrate is, 0.4568 grams

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