# basic chemistry formulas

• Jun 29th 2009, 08:32 AM
troubledmind
basic chemistry formulas
Having problems figurring out my math part of chemistry. Please someone help me.

How many moles of NaCl are dissolved in 395 mL of a 1.20M NaCl solution? i am not sure how i am to write this problem out.
• Jun 29th 2009, 09:09 AM
Raoh
hi(Hi)
In general we have : $n(moles) = c\times v$
here you have $C = 1.20 (mol/l)$ and $V = 0.395 (l)$
you do the math(Happy)
• Jun 29th 2009, 09:34 AM
alexmahone
$1.20 M = 1.20$ mol/litre
$395 ml=395*10^{-3}$ litre

Number of moles= $1.20*395*10^{-3}$=0.474
• Jun 29th 2009, 09:36 AM
troubledmind
omg
you have to be kidding. here i was trying so hard and it was just that easy.

you are the bomb. thank you
• Jun 29th 2009, 09:51 AM
troubledmind
i have several questions
how many moles of Al(NO3)3 would be contained in a 15.5 cc sample of a 0.250 M Al(NO3)3 solution> ? ok now will i need to get the atom mass of the element to use in this problem? will i use the mole 6.022 x1023?

how will it write it. i am online chemistr but lab 6 hrs on monday. only see teacher for those 6 hrsx. but busi with lab and no time for problem solving
(Happy)
• Jun 29th 2009, 10:22 AM
e^(i*pi)
Quote:

Originally Posted by troubledmind
how many moles of Al(NO3)3 would be contained in a 15.5 cc sample of a 0.250 M Al(NO3)3 solution> ? ok now will i need to get the atom mass of the element to use in this problem? will i use the mole 6.022 x1023?

how will it write it. i am online chemistr but lab 6 hrs on monday. only see teacher for those 6 hrsx. but busi with lab and no time for problem solving
(Happy)

You can use latex to represent chemical formulae here: $Al(NO_3)_3$. If you have any more chemistry questions there is also CHF (there is a link in my sig)

For this question you do not need to use Avogadro's number because it's asking for a value of moles rather than molecules. You do not need the atomic mass of $Al(NO_3)_3$ either since the question asks for the moles and the formula used give you a molar answer.

To solve this you may either use the method outlined in post 2 or you can use the formula:

$n = \frac{MV}{1000}$

Where
• $n$ is the number of moles
• $M$ is the molarity in $mol \: dm^{-3}$
• $V$ is the volume in $cm^3$

Plugging in the numbers gives: $n = \frac{0.250 \times 15.5}{1000} = 3.875\times 10^{-3} \text {{ mol }}$
• Jun 29th 2009, 03:26 PM
mr fantastic
Quote:

Originally Posted by troubledmind
Having problems figurring out my math part of chemistry. Please someone help me.