A 0.753 of CaCO is dissolved in HCl and diluted to a volume of 250.00ml what is

Question

A 0.753 of CaCO is dissolved in HCl and diluted to a volume of 250.00ml what is the molarity of calcium ions in the solution? o.53 X 100 15. B5000 F 3.31 2 Ethanol an has - enthalpy of evaporation of 386Jmol, and a normal boiling point of 18.4°C. What is the vapor of ethanol pressure at 25°C? R= 0.0083145 JK mol Pyap, T2A Hvap 11 T1 T2 Prap, H1 ANA ) O.CO314s Yas In A Hose (L-L 53.4 SC O.ODS3/4S Pop 50-4439943 An aqueous solution containing 22.5g of an unknown molecular compound in 100g of water had a freezing point of -2.5°C. Calculate the molar mass of the unknown compound. The freezing point depression constant for water is -1.86 kg/mole. Also, ATF Kem AT = 2.5-, Y 4. OS 2.5 x 4.5 x 100 4.65 a 3.5 T - 20. 7

Explanation / Answer

4)

Molar mass of CaCO3,

MM = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

= 1*40.08 + 1*12.01 + 3*16.0

= 100.09 g/mol

mass(CaCO3)= 0.753 g

number of mol of CaCO3,

n = mass of CaCO3/molar mass of CaCO3

=(0.753 g)/(100.09 g/mol)

= 7.523*10^-3 mol

volume , V = 250.00 mL

= 0.25 L

Molarity,

M = number of mol / volume in L

= 7.523*10^-3/0.25

= 3.01*10^-2 M

This is molarity of CaCO3

1 molecules of CaCO3 has 1 molecule of Ca2+ ion

So,

[Ca2+] = 3.01*10^-2 M

Answer: 3.01*10^-2 M

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