What is the concentration (M) of Ch3OH a solution prepared by dissolving of CH3OH sufficient water to give exactly 230 of solution?

1.59 M

Explanation:

What is the concentration (M) of CH₃OH a solution prepared by dissolving 11.7 g of CH₃OH sufficient water to give exactly 230 mL of solution?

Step 1: Given data

Step 2: Calculate the moles corresponding to 11.7 g of CH₃OH

The molar mass of CH₃OH is 32.04 g/mol.

11.7 g × 1 mol/32.04 g = 0.365 mol

Step 3: Calculate the molarity of the solution

M = moles of solute / liters of solution

M = 0.365 mol / 0.230 L = 1.59 M

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.