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You added 5.00 mL of 0.40 M NaOH in methanol to 20.00 mL of cooking oil. Calculate the number of moles of vegetable oil, methanol, and NaOH that are initially present in the sample. Assume the density of vegetable oil is 0.895 g/mL and the molar mass is 895 g/mol. Look up the density and molar mass of any other compounds as needed.

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jscob1234

Answer: 0.02 moles of vegetable oil, 0.002 moles of NaOH, 0.003 moles of methanol.

Explanation: First we need to know molarity M, is moles/liter. So solving for moles of vegetable oil first. We have 20mL vegetable oil and we want moles.

20.00mL • 0.895g/mL cancels out mL and gives us = 17.9g then we can divide by g/mol to get moles.

17.9g/895g/mol = 0.02moles of vegetable oil

Now with 5.00 mL of 0.40 M NaOH in methanol we solve for how many mL of each we have. First we'll change our mL to L.

5.00 mL • L/1000mL = 0.005L

Now 0.4 mol/L • 0.005L = 0.002 mol of  NaOH

This means we have 0.6 • 0.005L = 0.003 mol of methanol.

okpalawalter8

We have that for the Question it can be said that the number of moles of vegetable oil, methanol, and NaOH that are

  • M_C=0.03mol
  • M_m=0.124mol
  • Moles of NaOH=0.002

Generally the equation for the Moles  is mathematically given as

[tex]Moles=\frac{\rho*v}{molar mass}[/tex]

Therefore

For Cooking Oil

[tex]M_C=\frac{0.895*20}{895}\\\\M_C=0.03mol[/tex]

For Methanol

[tex]M_m=\frac{0.791*5}{32.04}[/tex]

M_m=0.124mol

For NaOH

[tex]Moles of NaOH=molarity *vol \\\\Moles of NaOH=0.40*(5*10^{-3})[/tex]

Moles of NaOH=0.002

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https://brainly.com/question/23379286

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