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Gom Sample - naci N NOOH + N Nazco, 0.1065 mol ' x 23.56X10L = NNOOH + NNa2CO nNaOH + nNascoz 0.002509 @ - from Sample-2 NHCl = nNaOH + 0.1065 mol LX 28.45x16L 2 * N Na2CO n naon + 2 NNa2CO3 + NNOOH --- 21 NzzCos () 0.0030 30 from ea 1) - equil NNa2CO3 nNaOH = 0.000521 mol 0.002509 0.00052) -0.001988 mol These amount are present in 25m2 of the total prepared 250 ml Solutim. so Amount present in the original sample (Amount present in a5 m2) x 10 - mol/ NaOH in = Oguginl Sample mass of 0.001988 mol X10 X 39.9979 0.795/40369 0.795 g 0.000521 mol X10 X 105.989 g mol 0.55220261 g 2 g mass of Na2CO3 in Oguginal sample 0.552