Question

A sample of chalk weighing 1.5 gms was dissolved in 200mlsolution required 50 ml of 0.2 N NaOH to neutralize the excess acidwhat is theweight of CaCO3 in the sample?​

Answer 1

Answer: Amount of HCl unreacted is determined by titration with NaOH Therefore, milliequivalent of acid left=molar equivalent of NaOH used for, milliequivalent of acid left =NV(NaOH)or, milliequivalent of acid left= 50×0.2=10 mileiq. Now,Total Molar Equivalent of acid= 200×0.1=20 mileiq.So, milliequivalent of acid reacted with CaCO3=20−10=10 mileiq. Now, mileiq. of CaCO3 present=mileiq. of acid reacted So, mileiq. of CaCO3=10 Molar Equivalent=Mass Equivalent mass×1000So, Mass of CaCO3=10×501000= 0.5 gNow, 1.5 g of chalk contains 0.5 g of CaCO3% of CaCO3=0.51.5×100=33.33%