from Sample - NACI nNaOH + N Na2CO 0.1065 mol L x 23.56 X703L nNaOH + NNa2CO3 O nNaOH + NNa2CO 0.002509 from Sample-2 NHOI nNaOH + 0.1065 mol LX 28.45X163L 2 * NNa2CO n naon + 2 N Na2CO3 NNUOH + 2n Nazcos 0.0030 30 foom eq (0) - equ NNa2CO3 nNaOH 0.000521 mol 0.002509 0.00052) = 0.001988 mol These amount are present in 25 m2 of the total prepoored 250 ml solutim. so Amount present in the original sample - (Amount present in 2 m2) x 10 - mass of NaOH in Original Sample 0.001988 mol X10 X 39.997g mol-'| 0.79514036 g 0.795 g mass of Na2CO3 in oguiginal simple - 0.00052) mol X10 X 105.989 g mol 0.55220261 g 0.552 g from Sample - NACI nNaOH + N Na2CO 0.1065 mol L x 23.56 X703L nNaOH + NNa2CO3 O nNaOH + NNa2CO 0.002509 from Sample-2 NHOI nNaOH + 0.1065 mol LX 28.45X163L 2 * NNa2CO n naon + 2 N Na2CO3 NNUOH + 2n Nazcos 0.0030 30 foom eq (0) - equ NNa2CO3 nNaOH 0.000521 mol 0.002509 0.00052) = 0.001988 mol These amount are present in 25 m2 of the total prepoored 250 ml solutim. so Amount present in the original sample - (Amount present in 2 m2) x 10 - mass of NaOH in Original Sample 0.001988 mol X10 X 39.997g mol-'| 0.79514036 g 0.795 g mass of Na2CO3 in oguiginal simple - 0.00052) mol X10 X 105.989 g mol 0.55220261 g 0.552 g