Question

Which solution contains the largest number of moles of chloride ions?

Answer 1

Answer : The correct option is, (e) 7.50 ml of 0.500 m $FeCl_3$

Explanation :

First we have to calculate the moles of the following options.

(a) $\text{Moles of }CaCl_2=\text{Molarity}\times \text{Volume}={0.100M}\times {0.030L}=0.003mole$

As, 1 mole of $CaCl_2$ contains 2 moles of chloride ion

So, 0.003 mole of $CaCl_2$ contains $0.003\times 2=0.006$ moles of chloride ion

The moles of chloride ion in $CaCl_2$ is, 0.006 mole

(b) $\text{Moles of }BaCl_2=\text{Molarity}\times \text{Volume}={0.500M}\times {0.010L}=0.005mole$

As, 1 mole of $BaCl_2$ contains 2 moles of chloride ion

So, 0.005 mole of $BaCl_2$ contains $0.005\times 2=0.01$ moles of chloride ion

The moles of chloride ion in $BaCl_2$ is, 0.01 mole

(c) $\text{Moles of }KCl=\text{Molarity}\times \text{Volume}={0.400M}\times {0.025L}=0.01mole$

As, 1 mole of $KCl$ contains 1 mole of chloride ion

So, 0.01 mole of $KCl$ contains $0.01\times 1=0.01$ moles of chloride ion

The moles of chloride ion in $KCl$ is, 0.01 mole

(d) $\text{Moles of }NaCl=\text{Molarity}\times \text{Volume}={1.000M}\times {0.004L}=0.004mole$

As, 1 mole of $NaCl$ contains 1 mole of chloride ion

So, 0.004 mole of $NaCl$ contains $0.004\times 1=0.004$ moles of chloride ion

The moles of chloride ion in $NaCl$ is, 0.004 mole

(e) $\text{Moles of }FeCl_3=\text{Molarity}\times \text{Volume}={0.500M}\times {0.0075L}=0.00375mole$

As, 1 mole of $FeCl_3$ contains 3 moles of chloride ion

So, 0.00375 mole of $FeCl_3$ contains $0.00375\times 3=0.01125$ moles of chloride ion

The moles of chloride ion in $FeCl_3$ is, 0.01125 mole

Hence, the largest number of moles of chloride ions present in (e) 7.50 ml of 0.500 m $FeCl_3$

Answer 2

Molarity = number of moles of solute/liters of solution
number of moles of solute = molarity x liters of solution

Part (a): 30.00 ml of 0.100m Cacl2
number of moles of CaCl2 =  0.1 x 0.03 = 3x10^-3 moles
1 mole of CaCl2 contains 2 moles of chlorine, therefore 3x10^-3 moles of CaCl2 contains 6x10^-3 moles of chlorine

Part (b): 10.0 ml of 0.500m bacl2
number of moles of BaCl2 = 0.5 x 0.01 = 5x10^-3 moles
1 mole of BaCl2 contains 2 moles of chlorine, therefore 5x10^-3 moles of BaCl2 contains 10x10^-3 moles of chlorine

Part (c): 4.00 ml of 1.000m nacl
number of moles of NaCl = 1 x 0.004 = 0.004 moles
1 mole of NaCl contains 1 mole of chlorine, therefore 4x10^-3 moles of NaCl contains 4x10^-3 moles of chlorine

Part (d): 7.50 ml of 0.500m fecl3
number of moles of FeCl3 = 0.5 x 0.0075 = 3.75x10^-3 moles
1 mole of FeCl3 contains 3 moles of chlorine, therefore 3.75x10^-3 moles of FeCl3 contains 0.01125 moles of chlorine

Based on the above calculations, the correct answer is (d)