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A student weighs out a 6.04 g sample of , transfers it to a 100. mL volumetric flask, adds enough water to dissolve it and then adds water to the 100. mL tick mark. What is the molarity of chromium(III) sulfate in the resulting solution

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Answer:

"0.1540 M" is the correct answer.

Explanation:

The given values are:

Mass of sample,

M = 6.04 g

Volume,

V = 100 mL

or,

  = 100×10⁻³ L

Molar weight of Chromium(III) sulfate

MW = 392.16 g/mol

Now,

The molarity will be:

= [tex]\frac{M}{MW\times V}[/tex]

By putting the values, we get

= [tex]\frac{6.04}{392.16\times 100\times 10^{-3}}[/tex]

= [tex]\frac{6.04}{39.216}[/tex]

= [tex]0.1540 \ M[/tex]

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