A solution is prepared by adding 0.100g of solid NaCl to 50.0mL of 0.100 M CaCl2

Question

A solution is prepared by adding 0.100g of solid NaCl to 50.0mL of 0.100 M CaCl2. What is the molarity of the chloride ion in the final solution? Assume that the volume of the final solution is 50.0 mL

Explanation / Answer

molarity = no. of moles/ volume in liters. first we find no.of moles of chloride ions. 0.1 M CaCl2 contains 0.1 moles of CaCl2 in 1 liter. ==> 0.2 moles of Cl- in 1 lt. in 50 ml, no. of moles = 0.2*(50/1000) = 0.01 moles. no. of moles of cl- in 0.1gm of NaCl = no. of moles of NaCl = 0.1/(23+35.5) = 1.709*10^(-3) = 0.0017 total no. of mole of Cl- = 0.01+0.0017 = 0.0117 molarity = 0.0117*1000/50 = 0.234 so molarity of chloride ions = 0.234 M

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