basic chemistry formulas

**troubledmind** · June 29th 2009, 07:32 AM

Having problems figurring out my math part of chemistry. Please someone help me.

stressed.

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.

**Raoh** · June 29th 2009, 08:09 AM

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

**alexmahone** · June 29th 2009, 08:34 AM

$1.20 M = 1.20$ mol/litre
$395 ml=395*10^{-3}$ litre

Number of moles= $1.20*395*10^{-3}$ =0.474

**troubledmind** · June 29th 2009, 08:36 AM

you have to be kidding. here i was trying so hard and it was just that easy.

you are the bomb. thank you

**troubledmind** · June 29th 2009, 08:51 AM

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

**e^(i*pi)** · June 29th 2009, 09:22 AM

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

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