Question

Unit 3 4, If we are to minimize the total number of chloride ions impacting the environment upon ap- plication of road salt in the winter, which would be better to use on the roadss.a 23 w/w% sodium chloride solution or a 23 w/w% calcium chloride solution? Show calculations to explain your choice! trying

Answer 1

Answer:

It would be better to use the NaCl solution.

Explanation:

We can calculate the amount of Cl⁻ in each salt.

NaCl

Taking into account the molar masses, there are 35.5 g of Cl⁻ every 58.5g of NaCl. So, for a 23 w/w% solution of NaCl:

$\frac{23gNaCl}{100gSolution} .\frac{35.5gCl^{-} }{58.5gNaCl} =\frac{14.0gCl^{-} }{100gSolution} = 14.0w/w \%$

CaCl₂

Taking into account the molar masses, there are 71.0 g of Cl⁻ every 111g of CaCl₂. So, for a 23 w/w% solution of CaCl₂:

$\frac{23gCaCl_{2}}{100gSolution} .\frac{71.0gCl^{-} }{111gCaCl_{2}} =\frac{14.7gCl^{-} }{100gSolution} = 14.7w/w \%$

The concentration of Cl⁻ in the NaCl solution is 14.0 w/w% and its concentration in the CaCl₂ solution is 14.7 w/w%, so we would choose the NaCl solution to minimize Cl⁻ concentration.