What is the concentration in molarity of a solution made using 30.0 grams of ch3oh in 300.0 ml of water?
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This solution has a molarity of 3.12 M = 3.12 mol/ L

Explanation:

Step 1: Data given

mass of CH3OH = 30.0 grams

Volume of the solution = 300.0 mL = 0.3 L

Molar mass of CH3OH (see periodic table)

⇒ C = 12 g/mol

⇒ H = 1.01 g/mol

⇒ O = 16 g/mol

  ⇒ 1*12 + 4*1.01 + 16 = 32.04 g/mol

Step 2: Calculate number of moles of CH3OH

Moles CH3OH = mass CH3OH / Molar mass CH3OH

Moles CH3OH = 30.0 grams / 32.04 g/mol

Moles CH3OH = 0.936 mol

Step 3: Calculate molarity of the solution

Molarity of solution = moles CH3OH / volume of water

Molarity solution = 0.936 mol /0.300 L

Molarity of the solution = 3.12 M

This solution has a molarity of 3.12 M = 3.12 mol/ L

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.

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