How many grams of ATP are needed to prepare 0.24 L of a 271.0 mM solution? The a

Question

How many grams of ATP are needed to prepare 0.24 L of a 271.0 mM solution? The anhydrous molecular weight of ATP is 478.2 g/mole, the ATP powder being used has 1.0 moles of water and 2.0 moles of Na+ per mole.

I understand that i'd have to multiply the volume and the M (converting it to M from mM) to get the moles of ATP. Then i was told to use the moles to multiply it with the molecular weight of ATP to obtain the grams of it. But its incorrect. So i was wondering if I'm missing an extra step or what.

Explanation / Answer

Volume of solution = 0.24 L

Molarity of solution = 271.0 mM

= 0.271 M

Moles of ATP = 0.24 * 0.271

= 0.06504

Molecular weight of anhydrous ATP = 478.2 g/mole

Mass of anhydrous ATP = 0.06504 * 478.2

= 31.10 g

1 mole ATP have 1.0 moles of water and 2.0 moles of Na+

Moles of water =  0.06504

Moles of Na+ ions = 2 * 0.06504 = 0.13008

Mass of water = 18 * 0.06504 = 1.17 g

Mass of Na+ ions = 23 * 0.13008 = 2.99 g

Mass of ATP needed = 31.10 + 1.17 + 2.99

= 35.26 g

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