A `100 mL` of tap water was titrated with `M//50 HCl` with methyle orange as indicator. If `30 mL of HCl` were required. Calculate the hardness of `CaCO_(3)` per `10^(3)` parts of water. The hardness is temporary.

Answer

`30 g`

Explanation

for temproray hardness :
`Ca(HCO_(3))_(2)+2HCl toCaCl_(2)+2CO_92)+2H_(2)O`
`2 mol of HCl -=1 mol of Ca(HCO_(3))_(2)`
`-= 1 mol of CaCO_(3)(100 parts)`
`1 mol` or `1` equivalent of `HCl-=50 g of CaCO_(3)`
`30 mL of M/50(=N/50)HCl`
`=30/50xx0.05 g of CaCO_(3)` in `100 mL of H_(2)O`
`-=0.03 g CaCO_(3) ` in `100 mL` of tap `H_(2)O`
`-=(0.03xx10^(5))/100=30 g of CaCO_(3)`