Question

50 ml HCl dissolve with 4g of CaCO3. After the reaction 1.5 g CaCO3 remain find the molality of HCl.
Reaction CaCo3+ HCl -----> CaCL2+ CO2+H2O​

Answer 1

Answer:

Using morality, Number of moles are calculated for 1000ml or 1L of the solution.

1000ml contains 0.75MHCl

HCl=0.75 mole.

Therefore, 25ml of 0.75M HCl will contains HCl=10000.75×25=0.01875mole

 2 moles of HCl reacts with =1mole of CaCO3

Therefore,

0.01875 mole of HCl will react with =21×0.01875

=0.009375mole

Molar mass of CaCO3=100g

Hence, the mass of 0.009375 of CaCO3=no.ofmoles×molarmass=0.9375g