If I have 340 mL of a 0.5 M $N a B r$ $NaBr$ solution, what will the concentration be if l add 560 mL more water to it?

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If I have 340 mL of a 0.5 M $N a B r$ $NaBr$ solution, what will the concentration be if l add 560 mL more water to it?

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Answer 1 · 2016-04-29T05:53:07

Approx. $0.20$ $0.20$ $m o l \cdot L^{- 1}$ $mol*L^-1$ with respect to $N a B r$ $NaBr$

Explanation:

$Concentration = \frac{Moles of Solute}{Volume of Solution}$ $"Concentration " = " Moles of Solute"/"Volume of Solution"$

$=$ $=$ $\frac{340 \times 10^{- 3} \cdot L \times 0.5 \cdot m o l \cdot L^{- 1}}{(340 + 560) \times 10^{- 3} L}$ $(340xx10^-3*Lxx0.5*mol*L^-1)/((340+560)xx10^-3L)$

The numerator gives the amount of $N a B r$ $NaBr$ in moles in the initial volume. This is divided again by the new volume of the solution $900 \times 10^{- 3} \cdot L$ $900xx10^-3*L$ .

$=$ $=$ $? ? \cdot m o l \cdot L^{- 1} ?$ $??*mol*L^-1?$

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If I have 340 mL of a 0.5 M $N a B r$ $NaBr$ solution, what will the concentration be if l add 560 mL more water to it?

1 Answer

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Humanities

... and beyond

If I have 340 mL of a 0.5 M NaBrNaBr solution, what will the concentration be if l add 560 mL more water to it?

1 Answer

Explanation:

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If I have 340 mL of a 0.5 M $N a B r$ $NaBr$ solution, what will the concentration be if l add 560 mL more water to it?