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Chemistry 15 Online
OpenStudy (anonymous):

The mass of a quantity of NiCl2 is 24.6 g. How many formula units are in the sample? A. 3.2 3 10^23 NiCl2 formula units B. 1.1 3 10^23 NiCl2 formula units C. 1.9 3 10^27 NiCl2 formula units D. 5.3 3 10^-21 NiCl2 formula units

OpenStudy (anonymous):

@JFraser Can you help?

OpenStudy (jfraser):

can you find the molar mass of \(NiCl_2\)?

OpenStudy (anonymous):

129.5994 g/mol correct?

OpenStudy (jfraser):

good. i'm going to round up to 130, just to make it faster

OpenStudy (jfraser):

ONE mole of \(NiCl_2\) will weigh 130g, that's what the molar mass tells us

OpenStudy (jfraser):

but you don't have 130g, you've only got 24.6g

OpenStudy (anonymous):

Okay.. So, what do I do next to get this? My mind has gone blank..

OpenStudy (jfraser):

one mole of anything is always the same \(number\) of things, right?

OpenStudy (anonymous):

Yeah?

OpenStudy (jfraser):

so \[1 mole \space NiCl_2 \space = \space 6.02*10^{23} \space units \space NiCl_2 = 130g\]

OpenStudy (anonymous):

So, would it be D?

OpenStudy (jfraser):

it's not D. you need to set up an equality of fractions

OpenStudy (jfraser):

\[\frac{6.02*10^{23} units}{130g} = \frac{?? units}{24.6g}\]

OpenStudy (anonymous):

Oh!! Okay, so set up that using the numbers in each option to see which one works best?

OpenStudy (jfraser):

no, solve the equality for the missing piece and compare it to the options

OpenStudy (anonymous):

So, would it be B?

OpenStudy (jfraser):

it is

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