What is the mass of nickel(ii) nitrate (182.71 g/mol) dissolved in 25.0 ml of 0.100 m ni(no3)2 solution?

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

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Alleei Alleei · Answer 2 · 2019-08-18T08:31:48+02:00

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