The specific rotation of (r)-2-butanol is −13.5°. what specific rotation would be observed for a 3: 1 mixture of (s): (r) 2-butanol?

If the specific rotation of (r)-2-butanol is −13.5° then the specific rotation of (s)-2-butanol is simply the opposite in sign 13.5°. In a 3:1 mixture of (s):(r) 2-butanol, 75% is (s) while 25% is r, so the specific rotation of the mixture is:

mixture rotation = 0.75 * (13.5°) + 0.25 * (- 13.5°)

mixture rotation = 6.75°

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