Molarity of the 250ml of standard ca solution

There is not sufficient/definite information as to how much CaCO3 was dissolved in how much water.

Following calculation assumes 0.3120 g of CaCO3 was dissolvd in 250 mL water.

Molecular mass of CaCO3 = 100.086
Number of moles = 0.3120 /100.086 = 0.0031173

If CaCO3 was dissolved in 1 L of water, then 
Molarity = 0.003117 M   (to 4 significant figures)

If the CaCO3 was dissolved in 250 mL of water, then molarity
= 0.0031173/250= 0.1247 M   (to 4 significant figures)

NOTE: there is a problem with the question.
The solubility of CaCO3 in water is 0.013 g/L, so 0.312 g of CaCO3 cannot dissolve in 1 L of water, much less in 250 mL.

Answer is: the equilibrium  shifts toward the reactants, to the left side of chemical reaction.it is because lead(ii) nitrate dissociates in water: pb(no₃)₂ → pb²⁺(aq) + 2no₃⁻(aq).according to lechatelier's principle, the position of equilibrium moves to counteract thechange (more lead cations, common ion), the position of equilibrium will move so that the concentration ofproducts decrease and concentration of reactants of chemical reaction increase.read more on -

  the volume of  0.220 m k2s    solution  required  to  completely  react with 175 ml  of 0.115m co(no3)2  calculate the moles  of co(no3)2 =  molarity x volume/1000175 x0.115/1000= 0.02  moleswrite  the reacting  equationk2s(aq)  +  co(no3)2  = 2 kno3(aq)  + cos(s)by  use  of mole ratio between k2s to co(no3)2 which  is  1: 1 the moles of k2s is therefore = 175  x0.115/1000= 0.02  molesvolume = moles/molarity=0.02/0.22=0.09  ml  of k2s