What is the molarity of a 50.0 ml aqueous solution containing 10.0 grams of acetic acid, ch3cooh?

3.34 M or 3.34 mol/L

Explanation:

The formula for molarity is moles of solution ÷ liters of solution. So, we have to convert grams into moles and mL to L.

To convert grams to moles, first find the molar mass of CH₃COOH by looking at the atomic mass of each individual element.

C - 2(12.01) = 24.02

H - 4(1.01) = 4.04

O - 2(16.00) = 32.00

= 60.06

Now divide the number of grams by the molar mass.

10.0 g CH₃COOH ÷ 60.06 g = 0.167 mol CH₃COOH

To convert milliliters to liters, divide by 1000.

50.0 mL ÷ 1000 = 0.05 L

Now we can calculate the molarity.

M = moles of solution ÷ liters of solution

= 0.167 mol ÷ 0.05 L

= 3.34 M or 3.34 mol/L

The molarity is 3.34 M.

Hope that helps.