A student prepares a weak acid solution by dissolving 0.2000 g HZ to give 100. mL solution. The titration requires 33.5 mL of 0.1025 M NaOH. Calculate the molar mass of the acid. WebAssign will check your answer for the correct number of significant figures. g/mol (a) Would the molar mass be too high, too low, or unaffected if the student accidentally used 0.1000 g in the calculation

a)M=0.20/(0.335*0.1025)= 0.20/ 0.034 = 5.88 g/mol

b) if 0.100g is used instead of 0.200g

M = 0.1 / 0.034 = 2.94  hence the molar mass will be too low

Explanation:

0.2000 gHZ gives  100ml acid solution

33.5 ml of 0.1025 M NaOH  is required to prepare it

the moles = mass / molar mass

mass = 0.200 gHZ

moles = 0.0335*100 * 0.1025 = 0.034

therefore molar mass = mass / moles

                                  M=0.20/(0.335*0.1025)= 0.20/ 0.034 = 5.88

if 0.100g is used instead of 0.200g

M = 0.1 / 0.034 = 2.94  hence the molar mass will be too low

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