Ask your own question, for FREE!
Chemistry 19 Online
OpenStudy (anonymous):

How many moles and how many grams of NaCl are present in 250cm^3 of a 0.500M NaCl solution?{Molar mass of NaCl = 58.44g}

OpenStudy (anonymous):

@mayankdevnani

OpenStudy (anonymous):

@mathslover

OpenStudy (mayankdevnani):

\[mean\] \[250cm ^{3}\]

OpenStudy (anonymous):

yeah

mathslover (mathslover):

hmn wait...

OpenStudy (anonymous):

ok

mathslover (mathslover):

ok so the unit of density is = ?

mathslover (mathslover):

just tell me this... so that i can proceed @KingAditya

mathslover (mathslover):

kg m^{-3}

OpenStudy (anonymous):

kg/m^3

mathslover (mathslover):

right...

mathslover (mathslover):

unit of volume = cm^3 ( CGS )

OpenStudy (anonymous):

yeah

mathslover (mathslover):

Ok so now... what does M denote ? Molar...?

OpenStudy (anonymous):

mole

OpenStudy (anonymous):

m waiting

OpenStudy (mayankdevnani):

m answering....

OpenStudy (mayankdevnani):

plz wait until i give you correct answer...

OpenStudy (anonymous):

ok @mayankdevnani

OpenStudy (anonymous):

can any one help me

OpenStudy (anonymous):

@ParthKohli

OpenStudy (australopithecus):

250cm^3 = mL

OpenStudy (australopithecus):

0.250L = 250mL \[Molarity = \frac{MolesOfSolute}{LitersOfSolution}\]

OpenStudy (australopithecus):

0.500M = x mole/0.250L 0.500M*0.250L = 0.125mol then use the following formula to solve for Grams: \[Moles = \frac{Grams}{MolecularMass}\]

mathslover (mathslover):

\[\large{0.125=\frac{grams}{58.44}}\] just multiply 0.125 * 58.44

OpenStudy (australopithecus):

yup yup

mathslover (mathslover):

though good work @Australopithecus ....

OpenStudy (anonymous):

thnx @australopithecus nd @mathslover

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!