How do I calculate the molarity of (NH 4 ) 2 Ce(NO 3 ) 6 using a given weight of

Question

How do I calculate the molarity of (NH4)2Ce(NO3)6 using a given weight of the sample, needed to prepare 100 ml of 0.040 M solution?

We're preparing a weight sufficient for 100 mL of 0.040 M solution => That weight is 2.2086g of (NH4)2Ce(NO3)6

Now they want the molarity of (NH4)2Ce(NO3)6 using this measured weight.

Here's my work:

2.2086 g / 548.26 g / 0.040 mol / ,1 L = 1M

or is it this?

2.2086 g / 548.26 g * 0.040 mol / 100 ml * 1000 ml = 0.001611 M

Are either of these correct? I think I'm doing it wrong.

Explanation / Answer

Molarity (M)=n/V

=(2.2086*1000)/(584*100) =3.7*10-2 M

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