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.

Cause its being chemically applied?