What is the concentration (m) of ch3oh in a solution prepared by dissolving 16.8 g of ch3oh in sufficient water to give exactly 230 ml of solution?

The Molarity concentration is expressed in units of moles / L. So let us first determine the number of moles of CH3OH, and then divide that amount by the total volume of 0.230 L of solution.

To determine the number of moles of CH3OH, divide the weight in grams of CH3OH by the molecular weight of CH3OH: (MW of CH3OH = 32 g / mol)

number of moles = 16.8 g / (32 g / mol)

number of moles = 0.525 mol CH3OH 

Then we calculate for molarity:

Molarity = 0.525 mol CH3OH / .230 L

Molarity = 2.2826 mol / L

Molarity = 2.28 M

take a better picture and ill do it

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%