Question

How many milliliters of a 0.595 MM NaOH NaOH solution are needed to completely saponify 18.5 gg of glyceryl tristearate (tristearin)

Answer 1

The value is
`x =104.63 \ mL`

Step-by-step explanation:

From the question we are told that

The concentration of NaOH is
`C = 0.595\ M`

The mass of glyceryl tristearate is
`m = 18.5 \ g`

The reaction of tristearate (tristearin) with NaOH is shown on the first uploaded image (Reference Wikidot)

From the reaction we see that 1 mole of tristearate (tristearin) (with molar mass Z_b = 891.5 g/mol ) reacted with 3 moles of NaOH (with molar mass Z_a = 40 g/mol)

So the mass of tristearate (tristearin) in the equation is (m_k = 1 *891.5 = 891.5g )

and

The mass of NaOH in the equation is (m_x = 3 * 40 = 120 g )

Generally the mass of NaOH that will react with
`m = 18.5 \ g` of glyceryl tristearate is mathematically represented as

`h = (18.5 * 120)/(891.5)`

=>
`h = 2.490 \ g`

The mass of NaOH in 0.595 M ( mol/L) (i.e 0.595 mol in L(1000 mL) )of NaOH solution is mathematically evaluated as

`w = Z_a * C`

=>
`w = 40 * 0.595`

=>
`w = 23.8 \ g`

So if

`w = 23.8 \ g` is in 1000mL of the given solution of NaOH

`h = 2.490 \ g` will be is x mL of the NaOH solution

So

`x = (2.490 * 1000 )/(23.8)`

`x =104.63 \  mL`