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How many ug of nickel (Ni) are required to make 25.00 nanoliters of a 1.25 mol/L solution? Be sure to report your answer to the correct number of significant figures.

Respuesta :

indrawboy

volume of Ni = 25 nL = 25 x 10⁻⁹ L

mol Ni = 25 x 10⁻⁹ L x 1.25 mol/L = 3.125 x 10⁻⁸

mass = mol x Ar Ni

mass = 3.125 x 10⁻⁸ x 59 g/mol

mass = 1.84 x 10⁻⁶ g = 1.84 μg

Answer:

Explanation:

"25.00 nanoliters of a 1.25 mol/L solution" contains

25.00*10^(-9) L * 1.25 mol/L

= 31.25*10^(-9) mol of Ni

Molar mass of Ni is 58.69 g/mol

So 31.25*10^(-9) mol of Ni

= 31.25*10^(-9) * 58.69 g

= 1834.06*10(-9) g

= 1.834*10(-6) g

= 1.834 ug of Ni

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