On chemical analysis, it is found that 200 mL of CaCl2 solution contains 3.01× 1022 chloride ions. Calculate the molarity of this solution.

Question

On chemical analysis, it is found that 200 mL of CaCl2 solution contains 3.01× 1022 chloride ions. Calculate the molarity of this solution. (A) O.125M (B) 0.178M (C) 0.21 OM (D) 0.253 M

Tardigrade Admin · Accepted Answer

underset1 textmol mathopCaCl2 xrightarrowCa2++ underset2 × 6.02 × 1023 textions mathop2Cl- ∴ 301× 1022Cl- ions will be present in CaCl2=(1× 3.01× 1022/2× 6.02× 1023) =0.025 mol Molarity of solution =(0.025 mol/0.200 L)=0.125 M

Q. On chemical analysis, it is found that 200 mL of $C a C l_{2}$ solution contains $3.01 \times 10^{22}$ chloride ions. Calculate the molarity of this solution.

O.125M

0.178M

0.21 OM

0.253 M

$1 mol C a C l_{2} C a^{2 +} + 2 \times 6.02 \times 10^{23} ions 2 C l^{-}$ $∴$ $301 \times 10^{22} C l^{-}$ ions will be present in $C a C l_{2} = \frac{1 \times 3.01 \times 10 ^{22}}{2 \times 6.02 \times 10 ^{23}}$ $= 0.025 m o l$ Molarity of solution $= \frac{0.025 m o l}{0.200 L} = 0.125 M$

Q. On chemical analysis, it is found that 200 mL of CaCl2​ solution contains 3.01×1022 chloride ions. Calculate the molarity of this solution.