You are given a 1.00 g sample of an unknown tri-protic acid, which you dissolve in 50.0 ml of water containing phenolphthalein indicator. you titrate the acid solution with standardized 0.400 m koh(aq). it requires 39.06 ml of the koh solution to produce a light pink indicator color. what is the molecular weight of the unknown acid?

21.33 g/mol

Explanation:

Considering:-

So, Given that:-

Let Molarity of tri-protic acid = x M

n-factor of tri-protic acid = 3 (3 dissociable H)

So,  

Normality of tri-protic acid = x/3 N

So, Given that:-

Let Molarity of KOH = 0.400 M

n-factor of KOH = 1 (1 dissociable OH)

So,  

Normality of KOH = 0.400 N

Considering:-

At equivalence point

Gram equivalents of acid = Gram equivalents of base

So,

Given  that:  

So,    

x = 0.93744 M

Also,

Considering:

Or,

Given :

For tri-protic acid :

Molarity = 0.93744 M

Volume = 50.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 50.0×10⁻³ L

Thus, moles of tri-protic acid :

Moles of tri-protic acid  = 0.046872 moles

Also, Given mass = 1.00 g

So,

Molar mass = Mass/Moles = 1.00g / 0.046872 moles = 21.33 g/mol

21.33 g/mol is the molecular weight of the unknown acid.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

i think it's c. if you look search electronic symbol on google and then click on wiki it shows you a