concentration of CH3OH in a solution prepared by dissolving 11.7 g CH3OH in sufficient water to give 217.0 mL?

Question

PLEASE HELP!!! Not homework - just trying to understand HOW TO DO THIS... Formula.


What is the concentration of CH3OH in a solution prepared by dissolving 11.7 g CH3OH in sufficient water to give 217.0 mL of solution?


(do i add carbon, hydrogen, and oxygen together? then divide by 11.7 grams to get (# x 10^#??) If so, then what?)

Answer 1

You're adding 11.7g of methanol to water to make a total volume of 217.0 ml. You need a couple of steps:

First, find the density of methanol - 0.791g / ml.
Second, find the volume of methanol added - 11.7 / 0.791 = 14.8ml
14.8ml in 217ml of total volume = 14.8 / 217.0 = 0.0682 * 100 =
6.82% methanol by volume.

By weight? 217.0 - 14.8 = 202.2ml or g of water. 202.2 + 11.7 = 219.9g total solution weight. 11.7 / 219.9 * 100 = 5.32% by weight.

Molar concentration? 11.7 / 32.04 (MW, MeOH) = 0.365mol in 217ml. = 1.68M.

Answer 2

No of mole of answer = MV the position M is concentration of answer and V is volume of answer. Molar mass of CH3OH = 32 g/mol No of mole of CH3OH = 11.7g / (32 g/mol) = 0.3656 mole 230 mL = 0.23L M = No of mole / V M = 0.3656 / 0.23 = a million.fifty 9 g/mol

Answer 3

this seeems quite easy

you have 11.7g CH3OH in 0.217L

so the concentration is 11.7/0.217 =53.91g/L

the other question is to find the molarity

the molecular weight of CH3OH is 32g for 1 mole

and here the molarity is 53.91/32= 1.69M