IF 250ML OF SOLUTION HAS 2.5GM OF NA2CO3

THEN 1000ML OF SOLUTION HAS (1000*2.5)/250 GM OF NA2CO3 =2500/250 =10GM

NOW THIS 10GM OF NA2CO3 IS PRSENT IN 1L OF SOLUTION AFTER IT HAS BEEN DILUTED

BECAUSE ONLY THE SOLUTION HAS BEEN DILUTED THE AMOUNT OF SOLUTE HAS NOT BEEN CHANGED.

1 MOL OF NA2CO3=106GM THEN 10GM HAS 10/106 MOL

MOLARITY =NO. OF MOLES OF SOLUTE / VOLUME oF SOLUTON IN LITRES =(10/106) /1L =10/106 M =0.094M