0.1447 g of a diprotic acid is dissolved in 20 ml water. The diprotic acid solution requires 11.74 ml of a 0.0909 m naoh titrant solution to reach the endpoint. Calculate the molar mass of the acid. Watch significant figures.

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Ответ:

queen9384

16.01.2020

Chemistry

Explanation:

Percentage ethylene by weight = 57%

Percentage propylene by weight = 43%

Suppose in 100 grams of polymer:

Mass of ethylene = 57 g

Mass of propylene = 43 g

Moles of ethylene = $\frac{57 g}{28 g/mol}=2.036 mol$

Moles of propylene = $\frac{43g}{42g/mol}=1.024 mol$

1 mole = $N_A =6.022\times 10^{23}$ molecules/ atoms

Units of ethylene = $2.036 mol\times N_A$

Units of propylene = $1.024 mol\times N_A$

a) Fraction of ethylene units:

$=\frac{2.036 mol\times N_A}{2.306 mol\times N_A+1.024 mol\times N_A}=\frac{509}{765}$

b ) Fraction of propylene units:

$=\frac{1.024mol\times N_A}{2.306 mol\times N_A+1.024 mol\times N_A}=\frac{256}{765}$

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