What NaCl concentration results when 204 mLof a 0.760 M NaCI solution is mixed with 417 mL of a 0.430 M NaCl solution?

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What NaCl concentration results when 204 mLof a 0.760 M NaCI solution is mixed with 417 mL of a 0.430 M NaCl solution?

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Answer 1 · 2016-08-30T06:04:01

$[N a C l]$ $[NaCl]$ $≅$ $~=$ $0.5 \cdot m o l \cdot L^{- 1}$ $0.5*mol*L^-1$

Explanation:

$Solution 1$ $"Solution 1"$ , $moles of NaCl$ $"moles of NaCl"$ $=$ $=$ $204 \times 10^{- 3} \cdot L \times 0.760 \cdot m o l \cdot L^{- 1} = 0.155 \cdot m o l$ $204xx10^-3*Lxx0.760*mol*L^-1=0.155*mol$ .

$Solution 2$ $"Solution 2"$ , $moles of NaCl$ $"moles of NaCl"$ $=$ $=$ $417 \times 10^{- 3} \cdot L \times 0.430 \cdot m o l \cdot L^{- 1} = 0.179 \cdot m o l$ $417xx10^-3*Lxx0.430*mol*L^-1=0.179*mol$ .

Since we must consider the volumes to be additive when these solutions are mixed, we have the following concentration with respect to $N a C l$ $NaCl$ .

$Concentration$ $"Concentration"$ $=$ $=$ $\frac{(0.155 + 0.179) \cdot m o l}{(204 + 417) \times 10^{- 3} \cdot L}$ $((0.155+0.179)*mol)/((204+417)xx10^-3*L)$

$≅$ $~=$ $0.5 \cdot m o l \cdot L^{- 1}$ $0.5*mol*L^-1$ with respect to $N a C l$ $NaCl$ .

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What NaCl concentration results when 204 mLof a 0.760 M NaCI solution is mixed with 417 mL of a 0.430 M NaCl solution?

1 Answer

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