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If water is added to 50 ml of a 0.04M solution so that it fills a 200 mL beaker, what is the final concentration?

1 Answer

Sep 28, 2016

Final concentration $=$ $=$ $0.01 \cdot m o l \cdot L^{- 1}$ $0.01*mol*L^-1$ .

Explanation:

$Concentration$ $"Concentration"$ $=$ $=$ $\frac{Moles of solute}{Volume of solution in litres}$ $"Moles of solute"/"Volume of solution in litres"$

$=$ $=$ $\frac{50 \times 10^{- 3} \cdot L \times 0.04 \cdot m o l \cdot L^{- 1}}{200 \times 10^{- 3} L}$ $(50xx10^-3*Lxx0.04*mol*L^-1)/(200xx10^-3L)$

$=$ $=$ $(50xxcancel(10^-3)*cancelLxx0.04*mol*L^-1)/(200xxcancel(10^-3L))$

$=$ $(1xxcancel(10^-3)*cancelLxx0.04*mol*L^-1)/(4xxcancel(10^-3L))$

$=$ $0.01*mol*L^-1$

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