Mole ConceptHard
Question
10 L of hard water required 0.56 g of lime (CaO) for removing hardness. Hence, temporary hardness in ppm (part per million 106) of CaCO3 is:
Options
A.100
B.200
C.10
D.20
Solution
Temporary hardness is due to HCO3- of Ca2+ and Mg2+
Ca(HCO3)2 +
+ H2O
mole of CaCO3 = 2
Ca(HCO3)2 +
+ H2Omole of CaCO3 = 2
ppm of CaCO3 = (2/104)×106 =200
Create a free account to view solution
View Solution FreeMore Mole Concept Questions
A mixture contains NaCl and unknown chloride, MCl. When 1 g of this mixture is dissolved in water and excess of AgNO3 so...Total number of electrons present in 4.4 g oxalate ion (C2O4 2–) is...A quantity of 0.2 g of an organic compound containing C, H and O on combustion yielded 0.147 g CO2 and 0.12 g water. The...The vapour density of gas A is four times that of B. If molecular mass of B is M, then the molecular mass of A is :-...A mixture of 254 g of iodine and 142 g of chlorine is made to react completely to give a mixture of ICl and ICl3. How ma...